Maharashtra Board Class 12 Physics Chapter 8 Electrostatics PDF Download

8. Electrostatics

Can You Recall?

What are conservative forces?

What is potential energy?

What is Gauss' law and what is a Gaussian surface?

8.1 Introduction

In XIth Std we have studied the Gauss' Law which gives the relationship between the electric charge and its electric field. It also provides equivalent methods for finding electric field intensity by relating values of the field at a closed surface and the total charges enclosed by that surface. It is a powerful tool which can be applied for the calculation of the electric field when it originates from charge distribution of sufficient symmetry. The law can be written as

\[\phi = \frac{q}{\varepsilon_0} = \oint \vec{E} \cdot d\vec{s}\]

where \(\phi\) is the total flux coming out of a closed surface and \(q\) is the total charge inside the closed surface.

8.2 Application of Gauss' Law

In this section we shall see how to obtain the electric field intensity for some symmetric charge configurations with the help of some examples.

Common Steps Involved In Calculating Electric Field Intensity By Using Gauss' Theorem

Describe the charge distribution (linear/surface/volume)

Obtain the flux by Gauss' theorem (Let this be Eq. (A))

Visualize a Gaussian surface and justify it.

With the electric field intensity E as unknown, obtain electric flux by calculation, using geometry of the structure and symmetry of the Gaussian surface (Let this be Eq. (B))

Equate RHS of Eq. (A) and Eq. (B) and calculate E.

8.2.1 Electric Field Intensity Due To Uniformly Charged Spherical Shell Or Hollow Sphere

Consider a sphere of radius R with its centre at O, charged to a uniform charge density \(\sigma\) (C/m²) placed in a dielectric medium of permittivity \(\varepsilon\) (\(\varepsilon = \varepsilon_0 k\)). The total charge on the sphere, \(q = \sigma \times 4\pi R^2\)

By Gauss' theorem, the net flux through a closed surface

\[\phi = \frac{q}{\varepsilon_0}\]

(for air/vacuum k=1) where q is the total charge inside the closed surface.

To find the electric field intensity at a point P, at a distance r from the centre of the charged sphere, imagine a concentric Gaussian sphere of radius r passing through P. Let ds be a small area around the point P on the Gaussian surface. Due to symmetry and spheres being concentric, the electric field at each point on the Gaussian surface has the same magnitude E and it is directed radially outward. Also, the angle between the direction of E and the normal to the surface of the sphere (ds) is zero i.e., \(\cos \theta = 1\)

\[\vec{E} \cdot d\vec{s} = E \, ds \, \cos\theta = E \, ds\]

flux \(d\phi\) through the area ds = E ds

Total electric flux through the Gaussian surface

\[\phi = \oint \vec{E} \cdot d\vec{s} = \oint E \, ds = E \oint ds\]

\[\therefore \phi = E \, 4\pi r^2\]

From equations (8.1) and (8.2),

\[\frac{q}{\varepsilon_0} = E \, 4\pi r^2\]

\[\therefore E = \frac{q}{4\pi\varepsilon_0 r^2}\]

Since \(q = \sigma \times 4\pi R^2\) We have \(E = \frac{\sigma \times 4\pi R^2}{4\pi\varepsilon_0 r^2}\)

\[\therefore E = \frac{\sigma R^2}{\varepsilon_0 r^2}\]

From Eqn. (8.3) it can be seen that, the electric field at a point outside the shell is the same as that due to a point charge. Thus it can be concluded that a uniformly charged sphere is equivalent to a point charge at its center.

Case (i) If point P lies on the surface of the charged sphere: r = R

\[\therefore E = \frac{q}{4\pi\varepsilon_0 R^2} = \frac{\sigma}{\varepsilon_0}\]

Case (ii) If point P lies inside the sphere: Since there are no charges inside \(\sigma = 0\),

\[\therefore E = 0\]

Teacher's Note

A uniformly charged sphere acts like a point charge when you are outside it. This is like how Earth's gravity affects objects above it - we use Earth's total mass at its center for calculations.

Exam Trick

Remember: Outside = point charge formula. Inside hollow = Zero field. Like Faraday cage - inside is always safe!

Points To Remember

Gauss' law relates electric field to charge using closed surfaces.
A uniformly charged sphere equals a point charge outside it.
Inside a hollow sphere, the electric field is always zero.
Gaussian surface must match the symmetry of the charge distribution.
Charges outside the Gaussian surface do not contribute to the flux.

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