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If work done in moving a charge of 20mC from infinity to a point O in an electric field is 15J,then what is the electric potential at this point.?
Most Upvoted Answer
If work done in moving a charge of 20mC from infinity to a point O in ...
We know, W = qV,
here, q = 20 × 10⁻³ C and W = 15J....
so,
V = 15/(20 × 10⁻³) = 750 V
Community Answer
If work done in moving a charge of 20mC from infinity to a point O in ...
Answer:

Explanation:

The electric potential at any point in an electric field is the amount of work done in bringing a unit positive charge from infinity to that point in the electric field.

We are given that the work done in moving a charge of 20mC from infinity to a point O in an electric field is 15J. We know that the work done in moving a charge q through a potential difference V is given by:

W = qV

Therefore, the electric potential at point O is given by:

V = W/q

Substituting the given values, we get:

V = 15J / 20mC

V = 750V

Therefore, the electric potential at point O is 750V.

Conclusion:

The electric potential at point O is 750V, which means that it takes 750J of work to bring a unit positive charge from infinity to point O in the electric field.
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If work done in moving a charge of 20mC from infinity to a point O in an electric field is 15J,then what is the electric potential at this point.?
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