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प्रश्न
Consider the situation of the previous problem. A charge of −2.0 × 10−4 C is moved from point A to point B. Find the change in electrical potential energy UB − UA for the cases (a), (b) and (c).
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उत्तर
Given:
Magnitude of charge, q = −2.0 × 10−4 C
(a) The electric field is along the x direction.
Thus, potential difference between (0, 0) and (4, 2),
dV = −E.dx = −20 × 4 = −80 V
Potential energy (UB − UA) between the points = dV × q
⇒ UB − UA = (−80) × (−2.0 × 10−4)
⇒ UB − UA = 160 × 10−4 = 0.016 J
(b) A = (4 m, 2m), B = (6 m, 5 m)
dV = −E.dx = − 20 × 2 = −40 V
Potential energy (UB − UA) between the points = dV × q
⇒ UB − UA = (−40) × (−2 × 10−4)
⇒ UB − UA = 80 × 10−4 = 0.008 J
(c) A = (0, 0) B = (6m, 5m)
dV = −E.dx = −20 × 6 = −120 V
Potential energy (UB − UA) between the points A and B = dV × q
⇒ UB − UA = (−120) × (−2 × 10−4)
⇒ UB − UA = 240 × 10−4 = 0.024 J
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