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 :
$\displaystyle (a) \frac{Q}{8\pi \epsilon_0 L} $
$\displaystyle (b) \frac{3Q}{8\pi \epsilon_0 L} $
$\displaystyle (c) \frac{3Q}{4 \pi \epsilon_0 L (ln2)} $
$\displaystyle (d) \frac{Q (ln2)}{4 \pi \epsilon_0 L} $
Click to See Answer :
Sol: Consider a small element of length dx of the rod at a distance x from O .
Charge on the element $\displaystyle dq = \frac{Q}{L}dx $
Potential at O due to this element is
$\displaystyle dV = \frac{1}{4\pi \epsilon_0} \frac{dq}{x} $
Potential at O due to entire rod is
$\displaystyle V = \int_{L}^{2L} \frac{1}{4\pi \epsilon_0} \frac{dq}{x} $
$\displaystyle V = \frac{Q}{4\pi \epsilon_0 L} \int_{L}^{2L} \frac{dx}{x} $
$\displaystyle V = \frac{Q}{4\pi \epsilon_0 L} [lnx]_{L}^{2L} $
$\displaystyle V = \frac{Q}{4\pi \epsilon_0 L} [ln 2L -lnL] $
$\displaystyle V = \frac{Q}{4\pi \epsilon_0 L} ln(\frac{2L}{L}) $
$\displaystyle V = \frac{Q (ln2)}{4 \pi \epsilon_0 L} $