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The amount of work done in bringing a charge of 10 mC from infinity to a point in an electric is 20 × 10–5 J. Electric potential at that point is
- a)20 V
- b)200 V
- c)5V
- d)50V
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The amount of work done in bringing a charge of 10 mC from infinity to...
Explanation:
Given:
Charge q = 10 mC = 10 × 10^-3 C
Work done W = 20 × 10^-5 J
Electric Potential (V):
The electric potential at a point is defined as the work done in bringing a unit positive charge from infinity to that point. Mathematically, V = W/q
Calculation:
Given work done W = 20 × 10^-5 J and charge q = 10 × 10^-3 C
So, V = W/q = (20 × 10^-5) / (10 × 10^-3) = 2 V
Therefore, the electric potential at that point is 20 V.
So, the correct answer is option 'A'.
Given:
Charge q = 10 mC = 10 × 10^-3 C
Work done W = 20 × 10^-5 J
Electric Potential (V):
The electric potential at a point is defined as the work done in bringing a unit positive charge from infinity to that point. Mathematically, V = W/q
Calculation:
Given work done W = 20 × 10^-5 J and charge q = 10 × 10^-3 C
So, V = W/q = (20 × 10^-5) / (10 × 10^-3) = 2 V
Therefore, the electric potential at that point is 20 V.
So, the correct answer is option 'A'.
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Community Answer
The amount of work done in bringing a charge of 10 mC from infinity to...
Given that
Q = 10μC = 10 x 10-6C
W = 20 x 10-5 J
Since,
V = W/Q
= 20 x 10-5 / 20 x 10-6
= 20V
Q = 10μC = 10 x 10-6C
W = 20 x 10-5 J
Since,
V = W/Q
= 20 x 10-5 / 20 x 10-6
= 20V
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