Answer:
1/9 E
Explanation:
The electric field produced by a charged sphere outside the sphere 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
In this problem, we have a sphere of radius R (smaller sphere) with a charge Q that produces an electric field of magnitude E at r=R.
For the larger sphere,
R' = 3R
Therefore, the electric field at r=R' will be
[tex]E'=k\frac{Q}{R'^2}=k\frac{Q}{(3R)^2}=\frac{1}{9}(k\frac{Q}{R^2})=\frac{E}{9}[/tex]
So, the electric field just outside the larger sphere will be 1/9 E.
