# Question #d4018

Jul 30, 2016

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

${E}_{Q} = \frac{k Q}{\frac{d}{2}} ^ 2 = \frac{4 k Q}{d} ^ 2$

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

${E}_{q} = - \frac{k q}{\frac{d}{2}} ^ 2 = - \frac{4 k q}{d} ^ 2$
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
$E = {E}_{Q} + {E}_{q} = \frac{4 k}{d} ^ 2 \left(Q - q\right)$

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

${E}_{q} = \frac{k q}{\frac{d}{2}} ^ 2 = \frac{4 k q}{d} ^ 2$

So net E will be

$E = {E}_{Q} + {E}_{q} = \frac{4 k}{d} ^ 2 \left(Q + q\right)$
And E will have same direction as${E}_{Q}$ i.e. away from +Q and towards -q.