Electrostatics 2 Question 15

16. A non-conducting ring of radius $0.5 \mathrm{~m}$ carries a total charge of $1.11 \times 10^{-10} \mathrm{C}$ 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

Show Answer

Answer:

Correct Answer: 16. (a)

Solution:

  1. $-\int_{l=\infty}^{l=0} \mathbf{E} \cdot \mathbf{d} \mathbf{l}=\int_{l=\infty}^{l=0} d V=V$ (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 $)=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}$

$$ =\frac{\left(9 \times 10^{9}\right)\left(1.11 \times 10^{-10}\right)}{0.5} \approx 2 \mathrm{~V} $$



NCERT Chapter Video Solution

Dual Pane