$(a)\;+\large\frac{q}{2}\qquad(b)\;-\large\frac{q}{2}\qquad(c)\;-\large\frac{qr}{2d}\qquad(d)\;-\large\frac{qd}{2r}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Answer : (c) $\;-\large\frac{qr}{2d}$

Explanation :

When a charged metallic sphere is touched with one of the sphere their densities will become same . Therefore charge on both the spheres becomes $\;\large\frac{q}{2}$

Now

After earthing the $\;1^{st}\;$ sphere let the charge on $\;1^{st}\;$ sphere becomes $\;q^{|}\;.$

Then

$\large\frac{k\;q^{|}}{r}+\large\frac{\large\frac{k\;q}{2}}{d}=0 \quad \; since\; \; d > > r$

$q^{|}=-\large\frac{q\;r}{2\;d}\;.$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...