CH24.Problems

Electric PotentialJH

Given: E and displacement, Required: work, Potential Diff

Work = displacement times force (if uniform; o.w. integrate): = q*E*d = 6e-13 J,

Potential difference: -w/q = -3.75e7 V

Suggested Problems

Answer: (a) 2.4x104 V/m = 24 kV/m ; (b) 2.9x103 V = 2.9 kV

Note: Electric potential is some alternative form of Electric field. However, it is a scalar unlike the vector E which make it easier to compute. You have to know how to go between E and V

Answer: (a) 30 V ; (b) 40 V ; (c) 5.5 m

Answer: (a) -4.5x103 V = -4.5 kV ; (b) same as (a)

Note: A and B are lying in equal potential surfaces. You can move in many paths between A and B and potential difference remains the same.

21. The ammonia molecule NH3 has a permanent electric dipole moment equal to 1.47 D, where 1 D = 1 debye unit = 3.34x10-30 C.m. Calculate the electric potential due to an ammonia molecule at a point 52.0 nm away along the axis of the dipole. (Set V = 0 at infinity.)

Answer: 1.63x10-5 V = 16.3 microV

Note: the potential from dipole drops like 1/r^2 and is proportional to the dipole moment rather than charge.

Answer: (a) 39 V/m ; (b) Toward plate 1

Note: Plate 1 and 2: he means from left to right as in the conventional direction starting always from left. Therefore, plate 1 is at left (-i).

Answer: (a) 0.225 J ; (b) aA = 45.0 m/s2 and aB = 22.5 m/s2 ; (c) vA = 7.75 m/s and vB = 3.87 m/s

Answer: (a) 1.2x104 V/m = 12 kV/m ; (b) 1.8x103 V = 1.8 kV ; (c) 0.058 m = 5.8 cm