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2. Electric Potential and Capacitance
hard
The electric potential at the centre of two concentric half rings of radii $R_1$ and $R_2$, having same linear charge density $\lambda$ is
A
$\frac{2 \lambda}{\epsilon_0}$
B
$\frac{\lambda}{2 \epsilon_0}$
C
$\frac{\lambda}{4 \epsilon_0}$
D
$\frac{\lambda}{\epsilon_0}$
(JEE MAIN-2023)
Solution
Potential at centre
$V=\frac{\left(\lambda \cdot \pi R_2\right)}{4 \pi \varepsilon_0 R_2}+\frac{\left(\lambda \cdot \pi R_1\right)}{4 \pi \varepsilon_0 R_1}$
$=\frac{\lambda}{2 \varepsilon_0}$
Standard 12
Physics
