Answer:
1/9 E0
Explanation:
The computation is shown below:
As we know that
[tex]E_0 = K \frac{Q}{r2}[/tex]
where,
E = Electric field strength
k = Coulomb's constant
Q = charge on the sphere
r = distance from the center of the sphere
It is given that
The radius of the larger sphere is three times larger than that of the smaller sphere i.e
[tex]R'=3R[/tex]
[tex]E'=k\frac{Q}{R'^2}[/tex]
[tex]=k\frac{Q}{(3R)^2}[/tex]
[tex]=\frac{1}{9}(k\frac{Q}{R^2})[/tex]
[tex]=\frac{E_0}{9}[/tex]
hence, the last option is correct
