Answer:
1.02*10^-5 C/m²
Explanation:
Given that
Radius of the smaller sphere, r = 0.05 m
Radius of the larger sphere, R = 0.12 m
Electric field, E = 2*10^5 V/m
Formula for the electric field is
E = Q/(4πεR²)
this then means that the surface charge density of the larger sphere is
Q/4πR² = Eε = 1.77*10^-6 C/m²
and
Q = 4πεER² = 3.203*10^-7 C
is the charge on the large sphere, which is the same as the charge on the small sphere since they are connected by the wire
so the surface charge density of the smaller sphere is
Q/4πr² = 4πεER²/4πr²
Q = εER²/r²
Q = 1.02*10^-5 C/m²