# An electric field of 20 N C–1 exists along the x-axis in space. Calculate the potential difference VB – VA where the points A and B are given by,(a) A = (0, 0); B = (4m, 2m)(b) A = (4m, 2m); B = (6m, 5m)(c) A = (0, 0); B = (6m, 5m)Do you find any relation between the answers of parts (a), (b) and (c)?

Given:
Magnitude of Electric field: E = 20 N C–1
E is along x-axis
Formula used:
As Electric field is along x-axis, potential difference will be along x-direction. Which means only x co-ordinates will be considered.
We know that,
Here dV is the change in potential : dV= VB-VA
E is the electric field along positive x axis and ds is the change in displacement.
(a)
A = (0, 0); B = (4m, 2m)
VB-VA= -20×(4-0) = -80 V
(b)
A = (4m, 2m); B = (6m, 5m)
VB-VA= -20×(6-4) = -40 V
(c)
A = (0, 0); B = (6m, 5m)
VB-VA= -20×(6-0) = -120 V
(d)
From (a),(b) and (c), we conclude that:
Potential difference of at points A = (0, 0), B = (6m, 5m)
= Potential difference at points A = (0, 0), B = (4m, 2m)
+ Potential difference at points A = (4m, 2m), B = (6m, 5m)

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