Presentation on theme: "CH24.Problems Electric Potential JH. Given: E and displacement, Required: work, Potential Diff Work = displacement times force (if uniform; o.w."— Presentation transcript:
CH24.Problems Electric Potential JH
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
Answer: (a) 2.4x10 4 V/m = 24 kV/m ; (b) 2.9x10 3 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) -4.5x10 3 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 NH 3 has a permanent electric dipole moment equal to 1.47 D, where 1 D = 1 debye unit = 3.34x 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) J ; (b) a A = 45.0 m/s 2 and a B = 22.5 m/s 2 ; (c) v A = 7.75 m/s and v B = 3.87 m/s
Answer: (a) 1.2x10 4 V/m = 12 kV/m ; (b) 1.8x10 3 V = 1.8 kV ; (c) m = 5.8 cm