Two concentric spherical conducting shells of radii R and 2R carry charges Q and 2Q respectively, change in electric potential on the outer shell when both are connected by a conducting wire $(k = \frac{1}{4 π ϵ _{0}})$ is:

Question

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Answer 1

The potential due to a charged spherical sphere with radius $r$ is given by

$V = \frac{1}{4 π ε _{o}} \frac{q}{r} = k \frac{q}{r}$

If we two spherical conducting shells of radii $R$ and $2 R$ carry charges $Q$ and $2 Q$ , then

$V_{1} = k \frac{Q}{2 R}$ + $k \frac{2 Q}{2 R} = k \frac{3 Q}{2 R}$

Charge remains constant before and after connecting them , so Final Charge $= 3 Q$

$V_{2} = k \frac{3 Q}{2 R}$

Hence $Δ V = V_{2} - V_{1} = 0$

Answer 2

The potential due to a charged spherical sphere with radius

r

is given by

V = \frac{1}{4 π ε _{o}} \frac{q}{r} = k \frac{q}{r}

If we two spherical conducting shells of radii

R

and

2 R

carry charges

Q

and

2 Q

, then

V_{1} = k \frac{Q}{2 R}

+

k \frac{2 Q}{2 R} = k \frac{3 Q}{2 R}

Charge remains constant before and after connecting them , so Final Charge

= 3 Q

V_{2} = k \frac{3 Q}{2 R}

Hence

Δ V = V_{2} - V_{1} = 0