17.1 Physics 6B Electric Field Examples Long

52
Physics 6B Electric Field Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Transcript of 17.1 Physics 6B Electric Field Examples Long

Page 1: 17.1 Physics 6B Electric Field Examples Long

Physics 6BElectric Field Examples

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Page 2: 17.1 Physics 6B Electric Field Examples Long

17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m.a) Find the magnitude and direction of the net electric field produced by q1 and

q2 at the origin.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.

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Page 3: 17.1 Physics 6B Electric Field Examples Long

q2 q1

x=0 x=0.2mx=-0.3mx

17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m.a) Find the magnitude and direction of the net electric field produced by q1 and

q2 at the origin.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.

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Page 4: 17.1 Physics 6B Electric Field Examples Long

q2 q1

x=0 x=0.2mx=-0.3m

The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

x

17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m.a) Find the magnitude and direction of the net electric field produced by q1 and

q2 at the origin.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.

2RkqE

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Page 5: 17.1 Physics 6B Electric Field Examples Long

q2 q1

x=0 x=0.2mx=-0.3m

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

At the origin, q1 will produce an E-field vector that points left, and q2 gives an E-field vector to the right.

E1 E2

x

17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m.a) Find the magnitude and direction of the net electric field produced by q1 and

q2 at the origin.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.

2RkqE

The electric field near a single point charge is given by the formula:

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Page 6: 17.1 Physics 6B Electric Field Examples Long

q2 q1

x=0 x=0.2mx=-0.3m

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

At the origin, q1 will produce an E-field vector that points left, and q2 gives an E-field vector to the right.

This is how we can put the +/- signs on the E-fields when we add them up.

E1 E2

x

17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m.a) Find the magnitude and direction of the net electric field produced by q1 and

q2 at the origin.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.

2RkqE

The electric field near a single point charge is given by the formula:

21total EEE

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Page 7: 17.1 Physics 6B Electric Field Examples Long

q2 q1

x=0 x=0.2mx=-0.3m

E1 E2

x

17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m.a) Find the magnitude and direction of the net electric field produced by q1 and

q2 at the origin.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.

2RkqE

The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

At the origin, q1 will produce an E-field vector that points left, and q2 gives an E-field vector to the right.

This is how we can put the +/- signs on the E-fields when we add them up.

21total EEE

CN

CN

2

9C

Nm9

2

9C

Nm9

total 500900)m3.0()C105)(109(

)m2.0()C104)(109(E 2

22

2

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q2 q1

x=0 x=0.2mx=-0.3m

Etotal

(This means 400 N/C in the negative x-direction)

x

17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m.a) Find the magnitude and direction of the net electric field produced by q1 and

q2 at the origin.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.

2RkqE

The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

At the origin, q1 will produce an E-field vector that points left, and q2 gives an E-field vector to the right.

This is how we can put the +/- signs on the E-fields when we add them up.

21total EEE

CN

CN

2

9C

Nm9

2

9C

Nm9

total 500900)m3.0()C105)(109(

)m2.0()C104)(109(E 2

22

2

CN

total 400E

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Page 9: 17.1 Physics 6B Electric Field Examples Long

q2 q3 q1

x=0 x=0.2mx=-0.3m

For part b) all we need to do is multiply the E-field from part a) times the new charge q3.

x

17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m.a) Find the magnitude and direction of the net electric field produced by q1 and

q2 at the origin.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.

2RkqE

The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

At the origin, q1 will produce an E-field vector that points left, and q2 gives an E-field vector to the right.

This is how we can put the +/- signs on the E-fields when we add them up.

21total EEE

CN

CN

2

9C

Nm9

2

9C

Nm9

total 500900)m3.0()C105)(109(

)m2.0()C104)(109(E 2

22

2

CN

total 400E (This means 400 N/C in the negative x-direction)

Etotal

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Page 10: 17.1 Physics 6B Electric Field Examples Long

q2 q3 q1

x=0 x=0.2mx=-0.3m

Note that this force is to the right, which is opposite the E-fieldThis is because q3 is a negative charge: E-fields are always set up as if there are positive charges.

x

17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m.a) Find the magnitude and direction of the net electric field produced by q1 and

q2 at the origin.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.

2RkqE

The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

At the origin, q1 will produce an E-field vector that points left, and q2 gives an E-field vector to the right.

This is how we can put the +/- signs on the E-fields when we add them up.

21total EEE

CN

CN

2

9C

Nm9

2

9C

Nm9

total 500900)m3.0()C105)(109(

)m2.0()C104)(109(E 2

22

2

CN

total 400E (This means 400 N/C in the negative x-direction)

For part b) all we need to do is multiply the E-field from part a) times the new charge q3.

N104.2)400)(C106.0(F 7CN9

onq3

Etotal

Fon3

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17.28 Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F?

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17.28 Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F?

The formula for electric force between 2 charges is 221

elec RqkqF

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Page 13: 17.1 Physics 6B Electric Field Examples Long

17.28 Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F?

The formula for electric force between 2 charges is

If both charges are doubled, we will have

221

elec RqkqF

221

221

elec Rqkq4R

)q2)(q2(kF

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Page 14: 17.1 Physics 6B Electric Field Examples Long

17.28 Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F?

The formula for electric force between 2 charges is

If both charges are doubled, we will have

So the new force is 4 times as large.

221

elec RqkqF

221

221

elec Rqkq4R

)q2)(q2(kF

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17.29 Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F?

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17.29 Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F?

The formula for electric force between 2 charges is

221

elec DqkqF

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Page 17: 17.1 Physics 6B Electric Field Examples Long

17.29 Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F?

The formula for electric force between 2 charges isWe want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance.

221

elec DqkqF

2new

212

21D

qkqD

qkq3

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17.29 Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F?

The formula for electric force between 2 charges is

Canceling and cross-multiplying, we get

We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance.

221

elec DqkqF

2new

212

21D

qkqD

qkq3

2312

new DD

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Page 19: 17.1 Physics 6B Electric Field Examples Long

17.29 Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F?

The formula for electric force between 2 charges is

Canceling and cross-multiplying, we get

We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance.

Square-roots of both sides gives us the answer:

221

elec DqkqF

2new

212

21D

qkqD

qkq3

2312

new DD

DD 31

new

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Page 20: 17.1 Physics 6B Electric Field Examples Long

17.30 When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released?

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Page 21: 17.1 Physics 6B Electric Field Examples Long

17.30 When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released?

Recall that Newton's 2nd law says that Fnet = ma.So this is really a problem about the force on the heavier charge.

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17.30 When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released?

Recall that Newton's 2nd law says that Fnet = ma.So this is really a problem about the force on the heavier charge.The formula for electric force between 2 charges is

221

elec dqkqF

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Page 23: 17.1 Physics 6B Electric Field Examples Long

17.30 When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released?

Recall that Newton's 2nd law says that Fnet = ma.So this is really a problem about the force on the heavier charge.The formula for electric force between 2 charges is

If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong:

221

elec dqkqF

221

51

2new

21

old51

new

dqkq

dqkq

FF

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17.30 When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released?

Recall that Newton's 2nd law says that Fnet = ma.So this is really a problem about the force on the heavier charge.The formula for electric force between 2 charges is

If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong:

We cancel common terms and cross-multiply to get

221

elec dqkqF

221

51

2new

21

old51

new

dqkq

dqkq

FF

22new d5d

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Page 25: 17.1 Physics 6B Electric Field Examples Long

17.30 When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released?

Recall that Newton's 2nd law says that Fnet = ma.So this is really a problem about the force on the heavier charge.The formula for electric force between 2 charges is

If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong:

We cancel common terms and cross-multiply to get

Square-root of both sides:

221

elec dqkqF

221

51

2new

21

old51

new

dqkq

dqkq

FF

22new d5d

d5dnew

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17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm

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17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm

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-4nC

x=0 x=0.8m

+6nCx

Page 28: 17.1 Physics 6B Electric Field Examples Long

-4nC

x=0 x=0.8m

+6nCx

The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm

2RkQE

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Page 29: 17.1 Physics 6B Electric Field Examples Long

a

For part a) which direction do the E-field vectors point?

17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm

-4nC

x=0 x=0.8m

+6nCx

The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

-4nC

x=0 x=0.8m

+6nCx

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2RkQE

Page 30: 17.1 Physics 6B Electric Field Examples Long

a

For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2 E1

E2

17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm

The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

-4nC

x=0 x=0.8m

+6nCx

Q1 = -4nC

x=0 x=0.8mx

Q2 = +6nC

21total EEE

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2RkQE

Page 31: 17.1 Physics 6B Electric Field Examples Long

a

For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2 E1

E2

17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm

The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

-4nC

x=0 x=0.8m

+6nCx

Q1 = -4nC

x=0 x=0.8mx

Q2 = +6nC

21total EEE

CN

2

9C

Nm9

2

9C

Nm9

total 1050)m6.0()C106)(109(

)m2.0()C104)(109(E 2

22

2

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2RkQE

Page 32: 17.1 Physics 6B Electric Field Examples Long

a

For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2 E1

E2

17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm

The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

-4nC

x=0 x=0.8m

+6nCx

Q1 = -4nC

x=0 x=0.8mx

Q2 = +6nC

21total EEE

CN

2

9C

Nm9

2

9C

Nm9

total 1050)m6.0()C106)(109(

)m2.0()C104)(109(E 2

22

2

For part b) E1 points left and E2 points right bE1

E2

Q1 = -4nC

x=0 x=0.8mx

Q2 = +6nC

21total EEE

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2RkQE

Page 33: 17.1 Physics 6B Electric Field Examples Long

a

For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2 E1

E2

17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm

The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

-4nC

x=0 x=0.8m

+6nCx

Q1 = -4nC

x=0 x=0.8mx

Q2 = +6nC

21total EEE

CN

2

9C

Nm9

2

9C

Nm9

total 1050)m6.0()C106)(109(

)m2.0()C104)(109(E 2

22

2

For part b) E1 points left and E2 points right bE1

E2

Q1 = -4nC

x=0 x=0.8mx

Q2 = +6nC

21total EEE

CN

2

9C

Nm9

2

9C

Nm9

total 5.312)m4.0()C106)(109(

)m2.1()C104)(109(E 2

22

2

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2RkQE

Page 34: 17.1 Physics 6B Electric Field Examples Long

a

For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2 E1

E2

17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm

The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

-4nC

x=0 x=0.8m

+6nCx

Q1 = -4nC

x=0 x=0.8mx

Q2 = +6nC

21total EEE

CN

2

9C

Nm9

2

9C

Nm9

total 1050)m6.0()C106)(109(

)m2.0()C104)(109(E 2

22

2

For part b) E1 points left and E2 points right bE1

E2

Q1 = -4nC

x=0 x=0.8mx

Q2 = +6nC

21total EEE

CN

2

9C

Nm9

2

9C

Nm9

total 5.312)m4.0()C106)(109(

)m2.1()C104)(109(E 2

22

2

For part b) E1 points right and E2 points leftc

E1

E2

Q1 = -4nC

x=0 x=0.8mx

Q2 = +6nC

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2RkQE

Page 35: 17.1 Physics 6B Electric Field Examples Long

a

For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2 E1

E2

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17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm

The electric field near a single point charge is given by the formula:

This is only the magnitude. The direction is away from a positive charge, and toward a negative one.

-4nC

x=0 x=0.8m

+6nCx

Q1 = -4nC

x=0 x=0.8mx

Q2 = +6nC

21total EEE

CN

2

9C

Nm9

2

9C

Nm9

total 1050)m6.0()C106)(109(

)m2.0()C104)(109(E 2

22

2

For part b) E1 points left and E2 points right bE1

E2

Q1 = -4nC

x=0 x=0.8mx

Q2 = +6nC

21total EEE

CN

2

9C

Nm9

2

9C

Nm9

total 5.312)m4.0()C106)(109(

)m2.1()C104)(109(E 2

22

2

For part b) E1 points right and E2 points leftc

E1

E2

Q1 = -4nC

x=0 x=0.8mx

Q2 = +6nC

21total EEE

CN

2

9C

Nm9

2

9C

Nm9

total 846)m0.1()C106)(109(

)m2.0()C104)(109(E 2

222

2RkQE

Page 36: 17.1 Physics 6B Electric Field Examples Long

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

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Page 37: 17.1 Physics 6B Electric Field Examples Long

x

y

Part a): TRY DRAWING THE E-FIELD VECTORS ON THE DIAGRAM 12

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For Campus Learning Assistance Services at UCSB

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

Page 38: 17.1 Physics 6B Electric Field Examples Long

x

y

Part a): both vectors point away from their charge.Since the distances and the charges are equal, the vectors cancel out.

Etotal = 0

12

E1 E2

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For Campus Learning Assistance Services at UCSB

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

Page 39: 17.1 Physics 6B Electric Field Examples Long

x

y

x

y

12

2 1

Part b): both vectors point away from their charge.

E1

E2

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For Campus Learning Assistance Services at UCSB

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

E1 E2Part a): both vectors point away from their charge.Since the distances and the charges are equal, the vectors cancel out.

Etotal = 0

Page 40: 17.1 Physics 6B Electric Field Examples Long

x

y

x

y

12

2 1

Positive x-direction

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For Campus Learning Assistance Services at UCSB

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

E1

E2

Part b): both vectors point away from their charge.

Part a): both vectors point away from their charge.Since the distances and the charges are equal, the vectors cancel out.

Etotal = 0

Positive x-direction

CN

2

9C

Nm9

1 2400)m15.0()C106)(109(E 2

2

CN

2

9C

Nm9

2 267)m45.0()C106)(109(E 2

2

E1 E2

Page 41: 17.1 Physics 6B Electric Field Examples Long

x

y

x

y

12

2 1

Positive x-direction

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For Campus Learning Assistance Services at UCSB

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

E1

E2

Part b): both vectors point away from their charge.

Part a): both vectors point away from their charge.Since the distances and the charges are equal, the vectors cancel out.

Etotal = 0

Positive x-direction

CN

2

9C

Nm9

1 2400)m15.0()C106)(109(E 2

2

CN

2

9C

Nm9

2 267)m45.0()C106)(109(E 2

2

CN

total 26672672400E

E1 E2

Page 42: 17.1 Physics 6B Electric Field Examples Long

x

yPart c): both vectors point away from their charge. We will need to use vector components to add them together.

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)

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For Campus Learning Assistance Services at UCSB

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

Page 43: 17.1 Physics 6B Electric Field Examples Long

x

y

12

E1,y

(0.15,- 0.4)

(0.15,0)(- 0.15,0)

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For Campus Learning Assistance Services at UCSB

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)Part c): both vectors point away from their charge. We will need to use vector components to add them together.

Page 44: 17.1 Physics 6B Electric Field Examples Long

x

y

12

E1,y

(0.15,- 0.4)

(0.15,0)(- 0.15,0)

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For Campus Learning Assistance Services at UCSB

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)Part c): both vectors point away from their charge. We will need to use vector components to add them together.

CN

2

9C

Nm9

1 5.337)m4.0()C106)(109(E 2

2

CN

y,1

CN

x,1

5.337E0E

Page 45: 17.1 Physics 6B Electric Field Examples Long

x

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)

E2

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17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)Part c): both vectors point away from their charge. We will need to use vector components to add them together.

E1,y

CN

2

9C

Nm9

1 5.337)m4.0()C106)(109(E 2

2

CN

y,1

CN

x,1

5.337E0E

Page 46: 17.1 Physics 6B Electric Field Examples Long

x

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)

The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.

0.4m

0.3m

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17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

E2

E1,y

Part c): both vectors point away from their charge. We will need to use vector components to add them together.

CN

2

9C

Nm9

2 216)m5.0()C106)(109(E 2

2

CN

2

9C

Nm9

1 5.337)m4.0()C106)(109(E 2

2

CN

y,1

CN

x,1

5.337E0E

Page 47: 17.1 Physics 6B Electric Field Examples Long

x

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)

The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.

0.4m

0.3m

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For Campus Learning Assistance Services at UCSB

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

E1,y

Part c): both vectors point away from their charge. We will need to use vector components to add them together.

CN

2

9C

Nm9

2 216)m5.0()C106)(109(E 2

2

E2,x

E2,y

CN

2

9C

Nm9

1 5.337)m4.0()C106)(109(E 2

2

CN

y,1

CN

x,1

5.337E0E

Page 48: 17.1 Physics 6B Electric Field Examples Long

x

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)

The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.

0.4m

0.3m

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For Campus Learning Assistance Services at UCSB

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

E1,y

Part c): both vectors point away from their charge. We will need to use vector components to add them together.

CN

2

9C

Nm9

2 216)m5.0()C106)(109(E 2

2

E2,x

E2,y

CN

2

9C

Nm9

1 5.337)m4.0()C106)(109(E 2

2

CN

y,1

CN

x,1

5.337E0E

CN

54

CN

y,2

CN

53

CN

x,2

8.172)()216(E

6.129)()216(E

Page 49: 17.1 Physics 6B Electric Field Examples Long

x

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)

The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.

0.4m

0.3m

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For Campus Learning Assistance Services at UCSB

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

E1,y

Part c): both vectors point away from their charge. We will need to use vector components to add them together.

CN

2

9C

Nm9

2 216)m5.0()C106)(109(E 2

2

E2,x

E2,y

CN

2

9C

Nm9

1 5.337)m4.0()C106)(109(E 2

2

CN

y,1

CN

x,1

5.337E0E

Add together the x-components and the y-components separately:

CN

CN

CN

y,total

CN

CN

CN

x,total

3.5108.1725.337E6.1296.1290E

CN

54

CN

y,2

CN

53

CN

x,2

8.172)()216(E

6.129)()216(E

Page 50: 17.1 Physics 6B Electric Field Examples Long

x

y

12

(0.15,- 0.4)

(0.15,0)(- 0.15,0)

The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)Part c): both vectors point away from their charge. We will need to use vector components to add them together.

CN

2

9C

Nm9

2 216)m5.0()C106)(109(E 2

2

CN

2

9C

Nm9

1 5.337)m4.0()C106)(109(E 2

2

CN

y,1

CN

x,1

5.337E0E

Add together the x-components and the y-components separately:

CN

CN

CN

y,total

CN

CN

CN

x,total

3.5108.1725.337E6.1296.1290E

Now find the magnitude and the angle using right triangle rules:

75.7º

Etotal

axisxbelow7.756.1293.510)tan(

5.526)3.510()6.129(E CN22

total

CN

54

CN

y,2

CN

53

CN

x,2

8.172)()216(E

6.129)()216(E

Page 51: 17.1 Physics 6B Electric Field Examples Long

x

yPart d): TRY THIS ONE ON YOUR OWN FIRST...

12(0.15,0)(- 0.15,0)

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17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

(0,0.2)

Page 52: 17.1 Physics 6B Electric Field Examples Long

x

yPart d): both vectors point away from their charge. We will need to use vector components to add them together.

12

E1 E2

(0,0.2)

(0.15,0)(- 0.15,0)

The 0.25m in this formula is the distance to each charge using the Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.

From symmetry, we can see that E2 will have the same components, except for +/- signs.

Now we can add the components (the x-component should cancel out)

The final answer should be 1382.4 N/C in the positive y-direction.

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For Campus Learning Assistance Services at UCSB

17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown.Find the magnitude and direction of the net electric field at:a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

CN

2

9C

Nm9

1 864)m25.0()C106)(109(E 2

2

CN

25.020.0

CN

y,1

CN

25.015.0

CN

x,1

2.691))(864(E4.518))(864(E

CN

25.020.0

CN

y,2

CN

25.015.0

CN

x,2

2.691))(864(E

4.518))(864(E

CN

CN

CN

y,total

CN

CN

CN

x,total

4.13822.6912.691E04.5184.518E