A 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

Question

A 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) (Q/8 π ε0 L) (B) (3Q/4 π ε0 L) (C) (Q/4 π ε0 L In 2) (D) (Q In 2/4 π ε0 L)

Tardigrade Admin · Accepted Answer

V=∫ limitsL2 L (k d Q/x)=∫ limitsL2 L (k((Q/L)) d x/x)=(Q/4 π ε0 L) ∫ limitsL2 L((1/x)) d x =(Q/4 π ε0 L)[ log e x]L2 L =(Q/4 π ε0 L)[ log e 2 L- log e L]=(Q/4 π ε0 L) In(2)

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

$\frac{Q}{8 π ε _{0} L}$

$\frac{3 Q}{4 π ε _{0} L}$

$\frac{Q}{4 π ε _{0} L I n 2}$

$\frac{Q I n 2}{4 π ε _{0} L}$

$V = L \int 2 L \frac{k d Q}{x} = L \int 2 L \frac{k ( \frac{Q}{L} ) d x}{x} = \frac{Q}{4 π ε _{0} L} L \int 2 L (\frac{1}{x}) d x$
$= \frac{Q}{4 π ε _{0} L} [lo g_{e} x]_{L}^{2 L}$
$= \frac{Q}{4 π ε _{0} L} [lo g_{e} 2 L - lo g_{e} L] = \frac{Q}{4 π ε _{0} L} I n (2)$

Q. A 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

8πε0​LQ​

4πε0​L3Q​

4πε0​LIn2Q​

4πε0​LQIn2​