Electrostatics 2 Question 15
16. A non-conducting ring of radius $0.5 m$ carries a total charge of $1.11 \times 10^{-10} C$ distributed non-uniformly on its circumference producing an electric field $E$ everywhere in space. The value of the integral $\int _{l=\infty}^{l=0}-\mathbf{E} \cdot \mathbf{d l}(l=0$ being centre of the ring) in volt is
(1997, 2M)
(a) +2
(b) -1
(c) -2
(d) zero
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Answer:
Correct Answer: 16. (c, d)
Solution:
- $-\int _{l=\infty}^{l=0} \mathbf{E} \cdot \mathbf{d} \mathbf{l}=\int _{l=\infty}^{l=0} d V=V$ (centre) $-V$ (infinity)
but $V$ (infinity) $=0$
$\therefore-\int _{l=\infty}^{l=0} \mathbf{E} \cdot \mathbf{d l}$ corresponds to potential at centre of ring.
and $\quad V($ centre $)=\frac{1}{4 \pi \varepsilon _0} \cdot \frac{q}{R}$
$$ =\frac{\left(9 \times 10^{9}\right)\left(1.11 \times 10^{-10}\right)}{0.5} \approx 2 V $$
