JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 14)
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 :
_24th_January_Evening_Shift_en_14_1.png)
$$\frac{\lambda}{2\in_0}$$
$$\frac{\lambda}{\in_0}$$
$$\frac{2\lambda}{\in_0}$$
$$\frac{\lambda}{4\in_0}$$
Explanation
$V_{1}=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{\lambda\left(\pi R_{1}\right)}{R_{1}}$
$$ \begin{aligned} V_{2}= & \frac{1}{4 \pi \varepsilon_{0}} \times \frac{\lambda\left(\pi R_{2}\right)}{R_{2}} \\\\ V_{\text {net }} & =V_{1}+V_{2} \\\\ & =2 \times \frac{1}{4 \pi \varepsilon_{0}} \pi \lambda \\\\ & =\frac{\lambda}{2 \varepsilon_{0}} \end{aligned} $$
$$ \begin{aligned} V_{2}= & \frac{1}{4 \pi \varepsilon_{0}} \times \frac{\lambda\left(\pi R_{2}\right)}{R_{2}} \\\\ V_{\text {net }} & =V_{1}+V_{2} \\\\ & =2 \times \frac{1}{4 \pi \varepsilon_{0}} \pi \lambda \\\\ & =\frac{\lambda}{2 \varepsilon_{0}} \end{aligned} $$
Comments (0)
