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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

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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)?  

संख्यात्मक
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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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अध्याय 29: Electric Field and Potential - Exercises [पृष्ठ १२३]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 29 Electric Field and Potential
Exercises | Q 56 | पृष्ठ १२३

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