Question:

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

Updated On: Jun 9, 2024
  • $\frac{\lambda}{4\epsilon_{0}}$
  • $\frac{\lambda}{8\epsilon_{0}}$
  • $\frac{\lambda}{12\epsilon_{0}}$
  • $\frac{\lambda}{16\epsilon_{0}}$
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The Correct Option is C

Solution and Explanation

Length of the arc =$r\theta =\frac{r\pi}{3} $
Charge on the are $=\frac{r\pi}{3} \times\lambda $
$\therefore $ Potential at centre $= \frac{kq}{r}$
$=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{r \pi}{3} \frac{\lambda}{r}=\frac{\lambda}{12 \varepsilon_{0}}$
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Concepts Used:

Electrostatic Potential

The electrostatic potential is also known as the electric field potential, electric potential, or potential drop is defined as “The amount of work that is done in order to move a unit charge from a reference point to a specific point inside the field without producing an acceleration.”

SI Unit of Electrostatic Potential:

SI unit of electrostatic potential - volt

Other units - statvolt

Symbol of electrostatic potential - V or φ

Dimensional formula - ML2T3I-1

Electric Potential Formula:

The electric potential energy of the system is given by the following formula:

U = 1/(4πεº) × [q1q2/d]

Where q1 and q2 are the two charges that are separated by the distance d.