Question

The charge on a sphere of radius r is +Q. At a point P which is outside this sphere and at a sufficient distance from it, the electric field is E. Now, another sphere of radius 2r and charge - 2Q is placed with P as the centre of this second sphere. Then, the electric field at the midpoint of the line joining the centres of the two spheres is :

• A. $E$
• B. $4E$
• C. $6E$
• D. $12E$

Answer 1

The correct option is D $12E$

If $OP=x$, according to question $OA=PA=x/2$.
By Gauss's law the field at P due to small sphere of charge Q is $E.4\pi x^2=\frac{Q}{{ϵ}_0}\Rightarrow E=\frac{Q}{4\pi {ϵ}_0{x}^2}$
As the spheres have charges $Q$ and $-2Q$ so net field at A is in the same direction means towards the big sphere.
here net field at A is $E_A=\frac{Q}{4\pi {ϵ}_0(x/2{)}^2}+\frac{2Q}{4\pi {ϵ}_0(x/2{)}^2}=\frac{12Q}{4\pi {ϵ}_0{x}^2}=12E$