An arc of radius r carries charge Q and the arc subtends a angle pie by 3 at the centre. What is the electric potential at the centre?
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Answer:
KQπ÷3
Explanation:
We know,
S = r(theta)
=) S = r π÷3
therefore total charge = rQπ ÷3
now potential (v) = KQ÷r
= K rQπ ÷ 3r
= KQπ÷3
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