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Question

A charge of 5 C experience a force of 5000N when it is kept in a uniform electric field. What is the potential difference between two points separated by a distance of 1 cm

A
10V
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B
250V
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C
1000V
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D
2500V
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Solution

The correct option is A 10V

Given,

Electrostatic Force F=qE ......(1)

Where,

q= charge

E=electric field

Potential difference V=Ed ...... (2)

Where,

E=Electric field

d=seperation

From (1) and (2)

V=Fdq=5000×0.015=10V

Hence, what is the potential difference between two points separated by a distance of 1cm is 10V


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