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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Physics » In The Case Of Solid Charged Sphere

In The Case Of Solid Charged Sphere

This article, in keeping with the field of electrostatics, covers the topic of solid charged spheres. The content covered includes the definition of the concept, and a few examples and the derivation of the relevant formula.

Table of Content
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Electrostatics is one of the essential branches in Physics, dealing with a charged particle in the static position. The electrostatic force was first accidentally discovered and investigated by the Dutch physicist Pieter Van Musschenbroek in 1976, from the University of Leiden. Following this, Charles Augustin de Coulomb, a French physicist, put forth the famous Coulomb Law, regarding electrostatics.

DEFINITION FOR ELECTROSTATICS:

Electrostatic force: The electrostatic force between particles due to electric charges is the attractive and repulsive force between particles. Electrostatics is a study of electromagnetic phenomena, present only when particles are static or unmoving; a phenomenon established only after the system attains a state of equilibrium.

This phenomenon is labelled as electrostatic because the charged particles will always be stationary or in rest.  That is why the current-carrying charged particles are known as electrostatics.

There are three basic laws in electrostatics. They state that similar charges repel each other, opposite charges attract each other, and charged objects attract neutral objects.

IN THE CASE OF SOLID CHARGED SPHERES:

To explain the special case of solid charged spheres the article will go on to explain important laws pertaining to the topic. These are Coulomb’s Law and Gauss’ Theorem.

Coulomb’s Law:

As proposed by Charles Augustin de Coulomb, this law states that the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges. Thus the corresponding formula is as follows:

Force = Product of the charges / distance2 

1 ⁄ 4πε0 * q1  q2  ⁄ r2 

where,  1 ⁄ 4πε0 = Constant

F = Force

q = charge

r = distance between the charges

This is also known as Coulomb’s constant, the S.I unit of which is  N m2 C-2

Note that a uniformly charged spherical shell interacts with external fields similarly to a point charge (as long as it stays uniformly charged). The material of the shell can be overlooked as long as the charge is spread in a thin, spherically symmetric shell.

The electric field of distributed charges, such as those produced by a uniformly charged spherical shell, cylinder, or plate, can be measured using Gauss’s law, which is as follows –

Gauss’s Theorem:

Gauss’s Theorem is one of the most important theorems used to calculate electric flux.

The total flux through a closed surface is equal to the  1 ⁄ ε times of the amount of charge enclosed by the surface.

The mathematical form of Gauss’s law is

Φ = Q ⁄ ε

Where,

Φ = electric flux

Q = total charge enclosed by the surface

ε= permittivity of the medium

The unit of permittivity is m-3 kg-1 S4 A2 

Gauss’ theorem assumes the surface to always be spherical.

The following are a few examples of solid charged spheres:

Assuming some quantities,

Φ = electric flux

Q = charge enclosed by the Gaussian surface

E = electric field

A = surface area

r = radius of the sphere

R = distance between the response point and the centre of the sphere

1 ⁄ 4πε0= electric field constant

The electric field of a uniformly conducting sphere:

If, r > R which means electric field outside the solid sphere

Φ = E. A

Φ = E . (4πr2) 

From Gauss’s theorem, we know that

Φ = Q ⁄ ε0

Comparing both the values of , Φ we get

E (4πr2) = Q ⁄ ε0

E= Q ⁄ 4πε0r2 

If, r < R which means electric field inside the solid sphere.

E= Qr ⁄ 4πε0R3 

When r < R the Gaussian surface encloses a lesser charge than the total charge in the sphere. So, the electric field will be less than the one outside the sphere.

Conclusion:

Electrostatics is defined as a study of the electromagnetic phenomena that occurs only if there are no moving charges. From Coulomb’s Law, it is proved that the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges. According to Gauss’ theorem, the total flux through a closed surface is equal to the times of the amount of charge enclosed by the surface.

In the case of solid charged sphere examples, the electric field inside the solid sphere is discussed.

faq

Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

What is the position of the external field at the surface of a metallic conductor?

Ans: As stated in the case of solid charged sphere examples, the external field at the surface of a metallic ...Read full

Give the formula of the electric field of a point charge using electric flux and Gauss’ law.

Ans: The electric flux Φ =EA where E is electric field, A is area...Read full

What does a uniformly charged surface consist of?

Ans: A uniformly charged surface consists of a constant charge per unit area over that surface. The electric ...Read full

Write the applications of Gauss’ law in everyday life.

Ans: As given in the case of solid charged sphere examples, the Gauss’ law is applied in everyday life in v...Read full

Ans: As stated in the case of solid charged sphere examples, the external field at the surface of a metallic conductor is perpendicular to the surface. The electric field lines are normal to their surface. The lines start and end on charges on the surface. But the electric field inside the conductor is zero.

Ans: The electric flux Φ =EA where E is electric field, A is area

On substituting the area:

Flux Φ =E × 4πr2..(1) where r is radius

According to Gauss’ law:

Electric flux Φ = Q ⁄ ε0….(2) where Q is charge and ε0 is electric constant

From equations (1) and (2):

Electric flux   E × 4πr2 =Q ⁄ ε0

Then the electric field of a point charge is: E= Q ⁄ 4πr2ε0 

Ans: A uniformly charged surface consists of a constant charge per unit area over that surface. The electric field inside a uniformly charged sphere is zero. In the case of two-dimensional surfaces, the amount of charge per unit area is called uniform surface charge density.

Ans: As given in the case of solid charged sphere examples, the Gauss’ law is applied in everyday life in various electrostatic problems.

  •   It is used to solve problems relating to spherical, planar, and cylindrical symmetries.
  •   It uses a simple method to calculate the electric field.
  •  It can be used to calculate the electric field of an infinite wire, infinite plane sheet, thin spherical shell, etc.

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