Answer :


Given:
Magnitude of Electric field: E = 20 N C–1
Magnitude of charge moved from A to B = –2.0 × 10–4 C
Formula used:
Change in Electrical potential energy is
Here, ΔU is change in Electrical potential energy, Δ V is change in potential and q is the charge displaced.
ΔU = UB-UA where UB is electric potential energy at B and UA is electric potential energy at A
(a)
A = (0, 0); B = (4m, 2m)
VB-VA= -20×(4-0) = -80 V
Thus Change in Electrical potential energy is
UB-UA = (VB – VA)× q
Δ U = -80 × -2.0× 10-4
Δ U = 0.016 J
(b)
A = (4m, 2m); B = (6m, 5m)
VB-VA= -20×(6-4) = -40 V
Thus Change in Electrical potential energy is
UB-UA = (VB – VA)× q
Δ U = -40 × -2.0× 10-4
Δ U = 0.008 J

(c)
A = (0, 0); B = (6m, 5m)
VB-VA= -20×(6-0) = -120 V
Thus Change in Electrical potential energy is
UB-UA = (VB – VA)× q
Δ U = -120 × -2.0× 10-4
Δ U = 0.024
Hence, for the cases (a),(b) and (c) the change in electric potential energy when the charge is moved from A to B is 0.016 J,0.008 J , 0.024 J respectively.


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