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The graph of radial component of E as a function of r will be:
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(A).
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(B).
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(C).
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(D).
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Last updated date: 23rd Feb 2024
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IVSAT 2024
Answer
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Hint: Figure contains two shells with inner radius and outer radii and the graphs depict a relationship between electric field due to the shells and the distance from the centre. The electric field depends on the charge enclosed and the distance from the centre of a shell. As the whole charge Is concentrated on the surface, the electric field inside the shell is zero.

Complete answer:
Electric field is the work done to bring a unit charge from infinity to a point in the field. Its SI unit is $C{{m}^{2}}$. It is given by-
$E=\dfrac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$ - (1)
Here, $E$ is the electric field
$q$ is the charge on the conductor
${{\varepsilon }_{0}}$ is the absolute permittivity
$r$ is the distance from the conductor
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The electric field inside a conductor is therefore, for $r < a$, E=0
When we draw a Gaussian surface with radius a, the charge enclosed inside the Gaussian surface is zero.
Therefore, from eq (1),
$\begin{align}
  & q=0 \\
 & \therefore E=0 \\
\end{align}$
Therefore, for r < a, $E=0$.
For a < r < b, there is no charge present inside a conducting shell as all the charge is concentrated on the surface. Thus, the electric field inside a conducting shell is zero.
Therefore, for a < r < b, $E=0$.
For b$E\propto \dfrac{1}{{{r}^{2}}}$
Therefore, for b < r < c, the electric field is inversely proportional to the square of distance from the centre
Again, for c < r < d, the electric field inside the shell is zero as there is no charge inside a conducting shell.
Therefore, for c < r < d, $E=0$
For dTherefore, for d < r, the electric field is inversely proportional to the radial component.
Therefore, the electric field is zero for r < a, a < r < b, c < r < d and is inversely proportional to the radial component for b < r < c, d < r.

Hence the correct graphical representation is (A).
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Note:
A Gaussian surface is an imaginary surface through which the electric lines of forces pass through normally. According to the Gauss law, the flux is equal to the charge enclosed in a Gaussian surface divided by absolute permittivity of free space. Question might arise for why is the electric field zero for r
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