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Question

Physics Question on Electrostatic potential

An arc of radius rr carries charge. The linear density of charge is λ\lambda and the arc subtends an angle π3\frac{\pi}{3} at the centre. What is electric potential at the centre?

A

λ4ϵ0\frac{\lambda}{4\epsilon_{0}}

B

λ8ϵ0\frac{\lambda}{8\epsilon_{0}}

C

λ12ϵ0\frac{\lambda}{12\epsilon_{0}}

D

λ16ϵ0\frac{\lambda}{16\epsilon_{0}}

Answer

λ12ϵ0\frac{\lambda}{12\epsilon_{0}}

Explanation

Solution

Length of the arc =rθ=rπ3r\theta =\frac{r\pi}{3}
Charge on the are =rπ3×λ=\frac{r\pi}{3} \times\lambda
\therefore Potential at centre =kqr= \frac{kq}{r}
=14πε0×rπ3λr=λ12ε0=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{r \pi}{3} \frac{\lambda}{r}=\frac{\lambda}{12 \varepsilon_{0}}