Question

Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity.

Answer 1

the potential at the center of curvature of the arc = v = Q ∕ (4πε∘a) or $\frac{kQ}{a}$

ATQ,

We have density of charge,

λ = $\frac{Q}{L}$

Where L is the rod's length, in this case the semicircle's length L = πr

Q is the charge on the rod

The potential created at the center by an differential element of charge is:

$k = \frac{dQ}{r}$

where k is the coulomb's constant

r is the distance from dQ to center of the circle

v = ∫ $\frac{K dQ}{a}$ , Where a = radius, k = 1 / 4πε∘

v =$\frac{kQ}{a}$ or Q ∕ (4πε∘a)

To learn more about potential from the given link

brainly.com/question/25923373

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