Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

The electric field at a distance of $\dfrac{{3R}}{2}$ from the centre of a charged conducting spherical shell of radius $R$ is $E$. The electric field at a distance $\dfrac{R}{2}$ from the centre of the sphere is:
(A) Zero
(B) $E$
(C) $\dfrac{E}{2}$
(D) $\dfrac{E}{3}$

seo-qna
Last updated date: 18th Jun 2024
Total views: 53.4k
Views today: 1.53k
Answer
VerifiedVerified
53.4k+ views
Hint: To solve this question, we need to use the property of the charge distribution within a hollow conductor. A spherical shell is basically a hollow sphere.

Complete step-by-step solution:
We know that a spherical shell is basically a sphere hollow from the inside. So we have been given a charged hollow sphere of radius $R$. Let the charge contained by the sphere be $Q$.
According to the question, the electric field at a distance of $\dfrac{{3R}}{2}$ from the centre of a charged conducting spherical shell is $E$.
The distance $d = \dfrac{{3R}}{2}$ can also be written as
$d = 1.5R$
So the given distance is greater than the radius of the spherical shell. We know that a charged spherical conductor behaves as a point charge concentrated at its centre, for all the distances greater than its radius. So the electric field can be given by
$E = \dfrac{Q}{{4\pi {\varepsilon _0}}}$
So the charge contained by the sphere can be given by
$Q = 4\pi {\varepsilon _0}E$
Therefore, the total charge contained by the sphere is equal to $4\pi {\varepsilon _0}E$.
Now we have to determine the electric field at a distance of $\dfrac{R}{2}$ from the centre of the spherical shell.
This distance can be given by
$d' = 0.5R$
So the given distance is less than the radius of the spherical shell. This means that the point situated at this distance is within the sphere.
Now, we know that when a hollow conductor is given a charge, then the charge gets distributed over the whole of its outer surface, to be at the maximum separation. So the whole charge of $4\pi {\varepsilon _0}E$ gets distributed over the outer surface of the spherical shell. This means that the charge within the given spherical shell will be equal to zero.
If we draw a Gaussian sphere concentric with the spherical shell, and pass through a point within the sphere, then the charge enclosed by this surface will be zero.
From the Gauss theorem we know that
$\oint {EdS} = \dfrac{q}{{{\varepsilon _0}}}$
Since $q = 0$ so we have
$\oint {EdS} = 0$
As the surface area of the sphere cannot be zero, so we have
$E = 0$
Therefore, the electric field at all points situated inside the spherical shell is equal to zero.
Thus, the electric field at a distance of $\dfrac{R}{2}$ from the centre of the spherical shell will be equal to zero.

Hence, the correct answer is option A.

Note: The information related to the electric field at the given distance, which is given in the question, is just extra information. We do not need this information for answering the given question.