Question

The charge per unit length of the four quadrant of the ring is $2\lambda ,-2\lambda ,\lambda$ and $-\lambda$ respectively. The electric field at the centre is

• A. 2
• B. 0
• C. 4
• D. 1

Answer 1

Answer: (B)

0

By symmetry, force will be horizontal
$2{\int }_0^{\frac{\pi }{2}}\frac{k\left(Rd\theta \right)}{R}\left(2\lambda \right)$
$=2\times 2R\lambda =4R\lambda \frac{k}{R}=4\lambda k$
Similarly $\to 2\lambda k$
Net field $=2\lambda k$
$=\left(0\right)$