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Question: A thin spherical shell of radius \(20\)cm is uniformly charged with \(Q\). \(V_0\) the potential on ...

A thin spherical shell of radius 2020cm is uniformly charged with QQ. V0V_0 the potential on its surface. Another point charge QQ is now placed at the centre of the shell. VV be the net potential (new) potential at a point 1515cm from it’s centre then, VV equals to:

Explanation

Solution

In order to this question, to calculate the new net potential of a thin spherical shell, first we will rewrite the given facts and then we will apply the formula of potential to show the relation between the old potential and the charge, and then we can find the new net potential with the same formula we applied.

Complete step by step answer:
The given radius of a thin spherical shell =20cm=20×10−2m = 20cm = 20 \times {10^{ - 2}}m
The given charge =Q = Q
V0{V_0} is the potential on the surface of the thin spherical shell.
net potential of the shell =V = V
Now, we will apply the formula of Potential:
V0=14π∈0Q20×10−2 ⇒V0×20×10−2=Q4π∈0  {V_0} = \dfrac{1}{{4\pi { \in _0}}}\dfrac{Q}{{20 \times {{10}^{ - 2}}}} \\\ \Rightarrow {V_0} \times 20 \times {10^{ - 2}} = \dfrac{Q}{{4\pi { \in _0}}} \\\ …………(i)
Now,
V1=14π∈0Q15×10−2 ⇒V1=Q4π∈0115×10−2  {V^1} = \dfrac{1}{{4\pi { \in _0}}}\dfrac{Q}{{15 \times {{10}^{ - 2}}}} \\\ \Rightarrow {V^1} = \dfrac{Q}{{4\pi { \in _0}}}\dfrac{1}{{15 \times {{10}^{ - 2}}}} \\\
So, according to the equation (i):
⇒V1=20×10−2×V015×10−2=43V0\Rightarrow {V^1} = \dfrac{{20 \times {{10}^{ - 2}} \times {V_0}}}{{15 \times {{10}^{ - 2}}}} = \dfrac{4}{3}{V_0}
Hence, the new net potential, V or V1V\,or\,{V^1} is equal to 43V0\dfrac{4}{3}{V_0}.

Note: At the surface, the value of the electric field shifts abruptly. We may infer that the electric field at point PP within the spherical shell is zero using the Gauss theorem. As a result, the Potential at point PP inside the spherical shell is obtained. It's all about the potential that the spherical shell provides.