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Byju's Answer

Standard XII

Physics

Corona Discharge

In the given ...

Question

In the given figure, $σ_{1}$ and $σ_{2}$ are the surface charge densities around points $1$ and $2$ respectively. Then:

σ_{1} > σ_{2}

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σ_{1} < σ_{2}

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σ_{1} = σ_{2}

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can't be interpreted

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Solution

The correct option is B $σ_{1} < σ_{2}$
We know that surface charge density $σ$ , of a conductor at a particular region on its surface is inversely proportional to the radius of curvature $r_{c}$ .
i.e. $σ \propto \frac{1}{r_{c}}$

In the given problem, since $(r_{c})_{1} > (r_{c})_{2}$
So, we can conclude that,
$σ_{2} > σ_{1}$

Hence, option (b) is the correct answer.

Why this question?

Note : For a sphere, $r_{c}$ is the same at all points, so $σ$ remains constant across the spherical surface.

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Similar questions

Q. In the given figure,

σ_{1}

and

σ_{2}

are the surface charge densities around points

1

and

2

respectively. Then the relation between the electric fields

E_{1}

and

E_{2}

near the points

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and

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respectively will be :

Q. The plates of a parallel plate capacitor are charged with surface densities

σ_{1}

and

σ_{2} (σ_{1} = σ

and

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respectively and placed horizontally. The electric field at points :

Q. Two parallel metallic plates have surface charge densities

σ_{1}

and

σ_{2}

as shown in figure. Match the following.

Table-1 Table-2

$(A)$ If $σ_{1} + σ_{2} = 0$	$(P)$ Electric field in region $III$ is towards right.
$(B)$ If $σ_{1} + σ_{2} > 0$	$(Q)$ Electric field in region $I$ is zero.
$(C)$ If $σ_{1} + σ_{2} < 0$	$(R)$ Electric field in region $I$ is towards right.
$(S)$ None.
$(T)$ Nothing can be said.

Two parallel plane sheets 1 and 2 carry uniform charge densities $σ_{1}$ and $σ_{2}$ (see Fig. 20.8). The electric field in the region marked I is $(σ_{1} > σ_{2})$ . (Consider the right direction as the positive direction)