A negative charge is placed at the midpoint between two fixed equal positive charges , separated by a distance 2d . If the negative charge is given a small distancement `x(xltltd)` perpendicular to the line joining the positive charges , how the force (F) developed on it will approximately depend on x ?

Options

A. `Fpropx`
B. `Fprop(1)/(x)`
C. `Fpropx^(2)`
D. `Fprop(1)/(x^(2))`

Answer

a

Explanation

Let us consider that , two positive charges be `+Q` and negative charge be -q
`AC=BC=r=(x^(2)+d^(2))^(1//2)` … (i)
Now , electric field at point C due to the charges +Q placed at A and B is
`E_("net")=2Ecostheta=2xx(kQ)/(r^(2))xx(x)/(r)=(2kQx)/(r^(3))`
`therefore` Force on charge - q will be , `F=-qE_(net)`
`=-(2kQqx)/(r^(3))=-(2kQqx)/((x^(2)+d^(2))^(3//2))` (using (i))
Now , for small displacement `(x),xltltd,x^(2)=0`
`thereforeF=-(2kQqx)/(d^(3))rArrFpropx`
So , the negative charge will oscillate along its mean position and the force acting on the negative charge will be restoring force.