Question

Suppose two equal charges of magnitude q each are kept one at x = 0 and the other at x = 2a, where can you place a third

charge so that the whole system is in equilibrium and how much charge has to be placed.​

Answer 1

Explanation:

Two identical point charges Q are kept at a distance r from each other. A third point charge is placed on the line joining these two charges such that all the three charges are in equilibrium. What is the magnitude, sign and position of the third charge?

Let the two charges +Q be placed at points A and B at a distance 'r'.

Let us suppose that the third charge 'q' is placed on the line joining the first and second charge such that AO=x and OB=r-x.

Net force on each of the three charges must be 0 for the system of charges to be in equilibrium.

If we assume that that 'q' is positive in nature then it will experience forces due to the other two charges in opposite direction and the net force on 'q' becomes 0. But, the force acting on Q at A or Q at B will not be 0. The forces will act in the same direction.

However, if charge 'q' is taken as negative then, on a charge Q forces due to other charges will act in opposite directions. Hence, the third charge must be negative in nature.

For charge -q to be in equilibrium, the force acting on '-q' due to +Q at A and +Q at B should be equal and opposite.

Now, using Coulomb's law,

14πε∘Qqx2=14πε∘qQ(r-x)2⇒ x2=(r-x)2⇒ x=± (r-x)⇒ x=r2

i.e., the position of the third charge is at x=r/2.

Now, to find the magnitude of the charge we will consider the case where charge +Q at A or at B are in equilibrium.

i.e., 14πε∘Qq(r/2)2 = 14πε∘ QQr2⇒ q=Q4

Mark me as Brainlist Answer. With your support I can got another rank.