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Question

A metal sphere with 1 m radius and a surface charge density of 10 coulombs/$m^2$ is enclosed in a cube of 10 m side. The total outward electric displacement flux normal to the surface of the cube is:

This question was previously asked in
CUET PG 2026 Agri-Business Management Question Paper (25-Mar-2026) (Shift 2)
The correct answer is

$40 \pi$ coulombs

Flux Calculation Using Gauss's Law

The problem asks for the total outward electric displacement flux normal to the surface of a cube that encloses a charged metal sphere. According to Gauss's Law for electric displacement, the total outward flux ($D_{flux}$) through a closed surface is equal to the total charge enclosed ($Q_{enc}$) within that surface.

The formula is given by: $ D_{flux} = \oint \vec{D} \cdot d\vec{A} = Q_{enc} $

Enclosed Charge Calculation

First, calculate the total charge ($Q_{enc}$) on the metal sphere. The sphere has a radius ($R$) of 1 m and a surface charge density ($\sigma$) of 10 C/$m^2$. The surface area ($A$) of the sphere is calculated as:

$ A = 4 \pi R^2 $

Substituting the given radius:

$ A = 4 \pi (1 \text{ m})^2 = 4 \pi \text{ m}^2 $

Now, calculate the total charge enclosed:

$ Q_{enc} = \sigma \times A $ $ Q_{enc} = (10 \text{ C/m}^2) \times (4 \pi \text{ m}^2) $ $ Q_{enc} = 40 \pi \text{ C} $

Determining Total Electric Displacement Flux

Since the cube encloses the charged sphere, the total outward electric displacement flux normal to the surface of the cube is equal to the total charge enclosed by the sphere.

$ D_{flux} = Q_{enc} $ $ D_{flux} = 40 \pi \text{ C} $

The side length of the cube (10 m) is irrelevant for calculating the total flux as long as it completely encloses the sphere.

Final Answer

The total outward electric displacement flux normal to the surface of the cube is $40 \pi$ coulombs.

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