Electric field due to a uniformly charged thin spherical shell

Case 1: At a point outside the spherical shell where r > R.

Let P be the point outside the shell at a distance r from the centre. Since the surface of the sphere is spherically symmetric, the charge is distributed uniformly throughout the surface.

A spherical Gaussian surface with the radius r and total charge enclosed on this Gaussian surface Q is selected. If Q > 0, then the electric field is radially pointed outward and if Q < 0, then the electric field is radially pointed inward.

From Gauss law, we know that

• \(\begin{array}{l}\oint \vec{E}.d\vec{A}=\frac{Q}{\epsilon _{0}}\end{array} \)

• \(\begin{array}{l}E\oint_{Gaussian\;surface} dA=\frac{Q}{\epsilon _{0}}\end{array} \)

• E.4πr² = Q/ε₀
• E = (1/4πε₀)(Q/r²)
• \(\begin{array}{l}\vec{E}=\frac{1}{4\pi \epsilon _0}\frac{Q}{r^{2}}\hat{r}\end{array} \)

We can say that the electric field at a point outside the shell will remain the same if the entire charge Q is concentrated at the centre of the spherical shell.

Case 2: At a point on the surface of a spherical shell where r = R.

Let P be the point at the surface of the shell at a distance r from the centre. In this case, r = R; since the surface of the sphere is spherically symmetric; the charge is distributed uniformly throughout the surface. From Gauss law, we know that

• \(\begin{array}{l}\oint \vec{E}.d\vec{A}=\frac{Q}{\epsilon _{0}}\end{array} \)

• \(\begin{array}{l}E\oint_{Gaussian\;surface} dA=\frac{Q}{\epsilon _{0}}\end{array} \)

• E.4πR² = Q/ε₀
• E = (1/4πε₀)(Q/R²)
• \(\begin{array}{l}\vec{E}=\frac{1}{4\pi \epsilon _0}\frac{Q}{R^{2}}\hat{r}\end{array} \)

Case 3: At a point inside the spherical shell where r < R.

Let P be the point inside the spherical shell at a distance r from the centre. In this case, r < R. From Gauss law, we know that

• \(\begin{array}{l}\oint \vec{E}.d\vec{A}=\frac{Q}{\epsilon _{0}}\end{array} \)

• \(\begin{array}{l}E\oint_{Gaussian\;surface} dA=\frac{Q}{\epsilon _{0}}\end{array} \)

• E.4πr₂ = Q/ε₀
• E = (1/4πε₀)(Q/r²)
• We know that the Gaussian surface does not enclose any charge, therefore, Q = 0.
• E = 0

Hence, we can say that the electric field due to the uniformly charged thin spherical shell is zero at all the points inside the shell.

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Frequently Asked Questions – FAQs

Q1

When do we use a pillbox?

A pillbox is used when the symmetry of the charge distribution is translational along the plane.

Q2

What are the three types of symmetry to be considered to determine the electric field?

The following are the three types of symmetry to be considered to determine the electric field:

• Planar symmetry
• Cylindrical symmetry
• Spherical symmetry

Q3

Is Gauss law valid for any closed and open surfaces?

The Gauss law is applicable for any closed surfaces and not for open surfaces.

Q4

What will be the charge inside the surface, if the flux of the electric field through a closed surface is zero?

The charge inside the surface will be zero if the flux of the electric field through a closed surface is zero.

Q5

What is the SI unit of electric flux?

The SI unit of electric flux is volt meters.

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