2. Electric Potential and Capacitance
medium

charge $Q$ is uniformly distributed over a long rod $AB$ of length $L$ as shown in the figure. The electric potential at the point $O$ lying at distance $L$ from the end $A$ is

A

$\frac{{Qln2}}{{4\pi {\varepsilon _0}L}}$

B

$\;\frac{Q}{{8\pi {\varepsilon _0}L}}$

C

$\;\frac{{3Q}}{{4\pi {\varepsilon _0}L}}$

D

$\;\frac{{3Q}}{{4\pi {\varepsilon _0}Lln2}}$

(JEE MAIN-2013)

Solution

Electric potential is given by,

$V=\int_{L}^{2 L} \frac{k d q}{x}=\int_{L}^{2 L} \frac{1}{4 \pi \varepsilon_{0}} \frac{\left(\frac{q}{L}\right) d x}{x}=\frac{q}{4 \pi \varepsilon_{0} L} \ln (2)$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.