Can you explain the answer of this question below:The electric field intensity due to a sphere (solid or hollow) at an external point varies as-A:1/rB:does not depend upon rC:1/r3D:1/r2The answer is d.

Question

Can you explain the answer of this question below:The electric field intensity due to a sphere (solid or hollow) at an external point varies as-&#8203;A:1/rB:does not depend upon rC:1/r3D:1/r2The answer is d.

Answer

It followw inverse square law as distance of observational point from source charge increases the electric field intensity decreases hence option d is correct

Answer

C

Answer

Outside the sphere,the formula for field is kq/r^2.so by this we can get the relation.

Answer

Electric Field Intensity due to a Sphere

The electric field intensity due to a sphere at an external point can be determined by applying Gauss's Law. According to Gauss's Law, the electric field flux through a closed surface is proportional to the enclosed charge.

Gauss's Law:
The electric field flux through a closed surface is given by the equation:

∮E⋅dA = Q/ε₀

Where,
- ∮E⋅dA represents the electric field flux through a closed surface,
- Q is the total charge enclosed by the surface,
- ε₀ is the permittivity of free space.

Electric Field Intensity of a Uniformly Charged Solid Sphere:
Consider a solid sphere of radius R with a uniformly distributed charge Q. To find the electric field intensity at an external point, we can imagine a Gaussian surface in the form of a concentric sphere with radius r (where r > R).

1. Charge Enclosed:
The charge enclosed by the Gaussian surface is the same as the total charge Q of the sphere since the entire charge is enclosed.

2. Electric Field Flux:
The electric field flux through the Gaussian surface is given by:

∮E⋅dA = Qenc/ε₀

3. Electric Field Intensity:
Since the electric field is radially symmetric, the electric field intensity E is constant in magnitude and direction on the Gaussian surface. Thus, the electric field flux can be expressed as:

E ∮dA = Qenc/ε₀

4. Gaussian Surface Area:
The surface area of the Gaussian surface is given by:

4πr²

5. Electric Field Intensity Equation:
Substituting the values into the previous equation, we get:

E * 4πr² = Q/ε₀

Simplifying the equation gives:

E = Q/(4πε₀r²)

Therefore, the electric field intensity due to a uniformly charged solid sphere decreases with the square of the distance from the center (1/r²).

Electric Field Intensity of a Hollow Sphere:
For a hollow sphere with a uniformly distributed charge, the electric field inside the hollow sphere is zero. This is because the electric field due to each infinitesimally small charge element cancels out with the electric field due to its opposite charge element on the opposite side of the sphere.

As a result, the electric field intensity at an external point due to a hollow sphere is the same as that due to a point charge located at the center of the sphere. The electric field intensity due to a point charge decreases with the square of the distance (1/r²).

Therefore, the electric field intensity due to a hollow sphere also decreases with the square of the distance from the center (1/r²).

Conclusion:
In conclusion, the electric field intensity due to a sphere (solid or hollow) at an external point varies as 1/r² (option D).

Answer

In case of a sphere (solid/hollow), electric intensity (E) at an external point varies as 1/r2, where r is the distance of the point from the centre of the sphere. In case of a line charge, E o 1/r and in case of an infinite plane sheet of charge, E does not depend upon r.

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