$\begin{array}{1 1}(A)\;Q \\(B)\;-Q\\(C)\;2Q \\(D)\;0 \end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

When C and A are placed in contact, charge of A equally divides in two spheres.

Therefore charge on each A and $C = +2Q$

Now, C is placed in contact with B, then charge on each B and becomes

$\qquad=\large\frac{2Q+(-10Q0)}{2}$$=-4Q$

When A and B are placed in contact then charge on each A and B becomes

$\qquad=\large\frac{2Q+(-4Q)}{2}$$=-Q$

Hence B is the correct answer .

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...