Consider the situation of the previous problem. A charge of – 2.0 × 10-4 C is moved from the point A to the point B. Find the change in electrical potential energy UB - UA for the cases (a), (b) and (c).

Question

Navjyot Kalra · Answer

Sol. (a) The Electric field is along x-direction Thus potential difference between (0, 0) and (4, 2) is, δV = –E × δx = – 20 × (40) = – 80 V Potential energy (U base B – U base A) between the points = δV × q = – 80 × (–2) × 10^–4 = 160 × 10^–4 = 0.016 J. (b) A = (4m, 2m) B = (6m, 5m) δV = – E × δx = – 20 × 2 = – 40 V Potential energy (UB – UA) between the points = δV × q = – 40 × (–2 × 10^–4) = 80 × 10^–4 = 0.008 J (c) A = (0, 0) B = (6m, 5m) δV = – E × δx = – 20 × 6 = – 120 V Potential energy (U base B – U base A) between the points A and B = δV × q = – 120 × (–2 × 10^–4) = 240 × 10^–4 = 0.024 J

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Consider the situation of the previous problem. A charge of – 2.0 × 10-4 C is moved from the point A to the point B. Find the change in electrical potential energy UB - UA for the cases (a), (b) and (c).

Consider the situation of the previous problem. A charge of – 2.0 × 10-4 C is moved from the point A to the point B. Find the change in electrical potential energy UB - UA for the cases (a), (b) and (c).

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