Electrostatics

We know that the electric field is zero inside a conductor if it is hollow or solid. That is,
EAinside = 0

On connecting them by a wire, both acquire a common potential, namely,
V0 =

Q+Q

C1+C2

=

2Q

4πε0RA+RB)

Therefore,
QA = C1V0 = (4πε0RA)V0 =

2QRA

RA+RB

QB = C2V0 = (4πε0RB)V0 =

2QRB

RA+RB

Since RA > RB, we get QA > QB. Now,

σA

σB

=

QA

RA2

×

RB2

QB

=

RA

RB

×

RB2

RA2

=

RB

RA

The electric field on the surface is

σ

ε0

. Since σA < σB. we get EAon‌surface < EBon‌surface