Physics Problems & Solutions: Electromagnetism

The long thin cylindrical glass rod shown above has length l and is insulated from its surroundings. The rod has an excess charge Q uniformly distributed along its length. Assume the electric potential to be zero at infinite distances from the rod. If k is the constant in Coulomb’s law, the electric potential at a point P along the axis of the rod and a distance l from one end is kQ/l multiplied by

A. 4/9
B. 1/2
C. 2/3
D. ln 2
E. 1

(GR8677 #98)

Solution:

Line Charge Density: $\lambda=\frac{q_{\text{enclosed}}}{l}$
$q_{\text{enc}} = Q$ , and for uniform line charge: $dQ=\lambda dl$

Electric Potential: $V=\frac{kQ}{r}$
$r=l$
$\\\rightarrow dV=\frac{kdQ}{l}=k\lambda \frac{dl}{l} \\\rightarrow V=k\lambda \int_{l}^{2l}\frac{dl}{l}=k\lambda \ln2=\frac{kQ}{l}\ln2$

Answer: D

Physics Problems & Solutions

Electromagnetism - Electric Potential

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