Answer:
1/9 E
Step-by-step explanation:
The electric field produced by a charged sphere outside the sphere is equal to that produced by a single point charge:
![E=k(Q)/(r^2)](https://img.qammunity.org/2020/formulas/physics/high-school/dlapwc2owjbl2muv9l2hawq453hgc50k7g.png)
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
![E'=k(Q)/(R'^2)=k(Q)/((3R)^2)=(1)/(9)(k(Q)/(R^2))=(E)/(9)](https://img.qammunity.org/2020/formulas/physics/high-school/uyq7dbxsef43lbohhjid66d6w2cp8fz2uv.png)
So, the electric field just outside the larger sphere will be 1/9 E.