Question

Question #d4018

Answer 1

we have 2 positive charges Q and q (Q>q) which are not equal, and are separated by distance d.

The electric field due to large charge Q at the mid point of distance of separation d will be given by

`E_Q=(kQ)/(d/2)^2=(4kQ)/d^2`

The electric field due to small charge q at the mid point of distance of separation d will be given by

`E_q=-(kq)/(d/2)^2=-(4kq)/d^2`
Negative sign indicates opposite direcion of the 2nd field w.r.to first one. k is the coulomb's constant.

So net field E at the mid point of their separation will be just the algebraic sum of their electric fields at a distance d/2.
Hence
`E=E_Q+E_q=(4k)/d^2(Q-q)`

And E will have same direction as`E_Q` i.e. away from Q and towards q.

If charges were now both negative the net direction will be just opposite.
If they were unlike charges +Q and -q then we are to add the E fields as they have same direction.

The electric field due to small charge -q at the mid point of distance of separation d will be given by

`E_q=(kq)/(d/2)^2=(4kq)/d^2`

So net E will be

`E=E_Q+E_q=(4k)/d^2(Q+q)`
And E will have same direction as`E_Q` i.e. away from +Q and towards -q.