Answer:
1/9 of that just outside the smaller sphere
Explanation:
The electric field strength produced by a charged sphere outside the sphere itself is equal to that produced by a single point charge:
[tex]E=k\frac{Q}{r^2}[/tex]
where
k is the Coulomb's constant
Q is the charge on the sphere
r is the distance from the centre of the sphere
Calling R the radius of the first sphere, the electric field just outide the surface of the first sphere is
[tex]E_0=k\frac{Q}{R^2}[/tex]
The second sphere has a radius which is 3 times that of the smaller sphere:
[tex]R'=3R[/tex]
So, the electric field just outside the second sphere is
[tex]E'=k\frac{Q}{R'^2}=k\frac{Q}{(3R)^2}=\frac{1}{9}(k\frac{Q}{R^2})=\frac{E_0}{9}[/tex]
So, the correct answer is 1/9.
