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प्रश्न
An electric field of 20 NC−1 exists along the x-axis in space. Calculate the potential difference VB − VA where the points A and B are
(a) A = (0, 0); B = (4 m, 2m)
(b) A = (4 m, 2 m); B = (6 m, 5 m)
(c) A = (0, 0); B = (6 m, 5 m)
Do you find any relation between the answers of parts (a), (b) and (c)?
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उत्तर
Given:
Electric field intensity, E = 20 N/C
The electric field is along the x-axis. So, while calculating the potential difference between points B and A using the formula VB − VA = E.ds, we will use the difference of the x-coordinates of these point as ds.
(a) A = (0, 0) B = (4 m, 2 m).
So, VB − VA = E.ds = 20 × (0 − 4 m) = − 80 V
(b) A = (4 m, 2 m), B = (6 m, 5 m)
⇒ VB − VA = E.ds = 20 × (4 − 6) = − 40 V.
(c) A = (0, 0), B = (6 m, 5 m)
⇒ VB − VA = E.ds = 20 × (0 − 6) = − 120 V.
Potential difference between points (0, 0) and (6 m, 5 m) = Potential difference between points (0, 0) and (4 m, 2 m) + Potential difference between points (4 m, 2 m) and (6 m, 5 m)
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