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Gauss Law And It's Application - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

• Gauss law, Applications of Gauss Law(I), Applications of Gauss Law(II), Applications of Gauss Law(IV), Applications of Gauss Law(V) are considered the most difficult concepts.

• 79 Questions around this concept.

Solve by difficulty

If the electric flux entering and leaving an enclosed surface respectively is $\phi _{1}$ and $\phi _{2}$, the electric charge inside the surface will be:

Let $\dpi{100} P(r)=\frac{Q}{\pi R^{4}}r$ be the charge density distribution for a solid sphere of radius $\dpi{100} R$ and total charge $\dpi{100} Q$ . For a point $\dpi{100} 'p'$ inside the sphere at a distance $\dpi{100} r_{1}$ from the centre of the sphere, the magnitude of the electric field is

Let there be a spherically symmetric charge distribution with charge density varying as $\dpi{100} \rho (r)=\rho _{0}\left ( \frac{5}{4}-\frac{r}{R} \right )\; upto\; r=R,and\; \rho (r)=0\; for\; r> R,$ where  $\dpi{100} r$ is the distance from the origin. The electric field at a distance $\dpi{100} r(r< R)$ from the origin is given by

Consider the charge configuration and spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface the electric field will be due to

A cone of base radius R and height h is  located in a uniform electric field parallel to its base. The electric flux entering the cone is :

Shown in the figure are two point charges +Q and -Q inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If is the surface charge on the inner surface and Q1 net charge on it and the surface charge on the outer surface and Q2 net charge on it then :

A thin spherical shell of radius R has  charge Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field produced by the shell in the range     where     is the distance from the centre of the shell?

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In a uniformly charged sphere of total charge Q and radius R the electric field E is plotted as a function of distance from the centre. The graph which would correspond to the above will be :

Concepts Covered - 6

Gauss law

Gauss's law

Gauss's law is one of the fundamental laws of physics which states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge. It is one of the four equations of Maxwell’s laws of electromagnetism.

The diagram below shows a locally uniform electric field E.  The lines are parallel and have constant density.  The same surface is inserted in three different orientations.  The maximum number of field lines is intercepted when unit vector normal to the surface, n, is parallel to the field E, while no field lines pass through the surface when n is perpendicular to the field.  In general, the number of field lines passing through an area  A is directly proportional to A*cosθ, where θ is the angle between the field direction and the unit vector n normal to the surface. This leads to the definition of the electric flux.

The surface on which Gauss's law is applied is called Gaussian Surface (as shown in the below figure).

Note- Remember that the closed surface in Gauss's law is imaginary. There need not be any material object at the position of the surface.

Gauss law for a closed surface states that :

$\phi _{\text{closed}}=\frac{q_{net}}{\epsilon_{0} }$  , $q_{net}$ is the total charge inside the closed surface. The closed surface on which we apply gauss law is called the gaussian surface.

Also, the flux can be written in the integral form as:

$\phi =\int \bar{E} \dot d\bar{S}$   S is the area enclosed and E is the electric field intensity passing through it.

The usefulness of Gauss's law :

1. Gauss's law is useful when the gaussian surface has symmetry about the charge.

2. We can take any gaussian surface but the Gaussian surface should not pass through the charge. iI can pass through the charge but it can pass through continuous charge distribution.

The limitation of Gauss law-

Gauss law is applicable to certain symmetrical shapes it cannot be used for disk and ring and an electric dipole etc.

Applications of Gauss Law(I)

Applications of Gauss Law-I:

Electric field due to infinite linear charge :

The electric field of an infinite line charge with a uniform linear charge density can be obtained by using Gauss's law. Considering a Gaussian surface in the form of a cylinder at radius r, the electric field has the same magnitude at every point of the cylinder and is directed outward. The electric flux is then just the electric field times the area of the cylinder.

Earlier we have derived the electric field due to the infinite line charge as :

$E= \frac{\lambda}{2 \pi \epsilon_{0} r}$ where $\lambda$ is linear charge density. Here the gaussian surface will be the cylinder around the linear charge of length l .

Gauss's law :

$\phi =\oint E \cdot dA = \frac{Q_{inside}}{\epsilon_{0}}$

Here cylinder has 3 surfaces: 1. Upper 2. lower   3. Curved

$\phi _{cylinder}=\phi _{1}+\phi _{2}+\phi _{3}$

$=0+0+\int EdA \cos{0}$

$\implies \phi =\int EdA =\frac{Q_{inside}}{\epsilon_{0}}$

$\implies \phi =E \int dA =\frac{Q_{inside}}{\epsilon_{0}}$

$\implies \phi =E (2\pi r l)=\frac{Q_{inside}}{\epsilon_{0}}$

$\implies E = (2\pi r l)=\frac{\lambda l}{2 \pi a \times l\times \epsilon_{0}}$   since, $\lambda= \frac{Q_{inside}}{l}$

$\implies E=\frac{\lambda}{2\pi\epsilon r}$

Therefore E is inversely proportional to r.

Applications of Gauss Law(II)

Applications of Gauss Law(II)

Electric field in a conductor:

We have studied that conductors have a large number of free electrons that are free to move inside the conductors but metal ions are fixed. Now if we place this conductor in an electric field, due to the electric field, electrons will experience a force.

The total electric field at any point in the conductor is the vector sum of the original electric field and the electric field due to the redistributed charged particles. Since they are oppositely directed, the two contributions to the electric field inside the conductor tend to cancel each other. Now comes the profound part of the argument: the two contributions to the electric field at any point in the conductor exactly cancel. We know they have to completely cancel because, if they didn’t, the free-to-move-charge in the conductor would move as a result of the force exerted on it by the electric field. And the force on the charge is always in a direction that causes the charge to be redistributed to positions in which it will create its own electric field that tends to cancel the electric field that caused the charge to move. The point is that the charge will not stop responding to the electric field until the net electric field at every point in the conductor is zero.

The electric field is zero at all points inside the conductor, and, while the total charge is still zero, the charge has been redistributed as in the following diagram:

These charges which are appearing on the surface are called induced charges. Due to these induced charges, the electric field will be produced and that is the induced electric field.

The direction of this electric field will be from positive to negative.

$E_{net}= E_{in}+ E_{ext}$

and $E_{net}= 0$ inside the conductor.

now if we make a gaussian surface inside the conductor

We know that Einside =0

Therefore, $\phi =\oint E \cdot ds=\frac{q_{inside}}{\epsilon _{0}}=0$

Hence, $q_{inside}=0$ charge inside the conductor zero.

Applications of Gauss Law(III)

Applications of Gauss Law(III):

Electric field due to cylinders :

1. Solid conducting/ hollow conducting

The electric field of an infinite cylindrical conductor with a uniform linear charge density can be obtained by using gauss' law. Considering a  Gaussian surface in the form of a cylinder at radius r > R, the electric field has the same magnitude at every point of the cylinder and is directed outward. The electric flux is then just the electric field times the area of the cylinder.

when the charge is on the surface we need to take account of the surface charge density.

$\sigma =\frac{\text{charge}}{\text{area}}$

For a uniformly charged cylinder of radius, R. We have to consider these three areas where

a) Inside the cylinder (r<R)

b)  At the surface (r=R)

c) outside the cylinder (r>R).

(a). Inside  (r<R)

E =0

(b). Outside (r>R)

$E=\frac{\sigma R}{\epsilon _{0}r}$   or  $\dpi{100} E =\frac{\lambda }{2\pi \varepsilon _{0}r}$

(c). At the surface (r=R)

$E=\frac{\sigma }{\epsilon _{0}}$    or $\dpi{100} E =\frac{\lambda }{2\pi \varepsilon _{0}R}$

2. Solid non-conducting

In the case of a solid non-conducting cylinder, the charge is not only on the surface but also distributed through the whole volume. Therefore, volume charge density, $\rho = \frac{\text{charge}}{\text{volume}}$

(a). Inside  (r<R)

$E = \frac{\rho r}{2\epsilon _{0}}$  or $\dpi{100} E =\frac{\lambda r}{2\pi \varepsilon _{0}R^{2}}$

(b). Outside (r>R)

$E=\frac{\rho R^2}{2\epsilon _{0}r}$

(c). At the surface (r=R)

$E=\frac{\rho R }{2\epsilon _{0}}(\text{max.})$

Applications of Gauss Law(IV)

FIeld of an infinite plane sheet -

It is one of the very important case and we are going to find the electric field of an infinte plane sheet with the help of Gauss's law -

Let us consider a thin, flat, infinite sheet which consists of uniform positive charge per unit area $\sigma$.  We can see that there is symmetry in this lamina. So, to take advantage of these symmetry properties, we use a cylinder as our Gaussian surface whose axis is perpendicular to the sheet of charge, with ends of area A .

We can also observe that the charged sheet passes through the middle of the cylinder's length and because of this flux through each end is EA. This is because $\vec{E}$ is perpendicular to the charged sheet and parallel to the area vector of the flat face. The $\vec{E}$ is along the curved surface i.e perpendicular to the area vector so, the flux will be zero through this. Then the total flux will be 2EA. Now the net charge within the Gaussian surface can be calculated as -

$Q_{enclosed} = \sigma A$

So we can write that by Gauss's law -

$2EA = \frac{\sigma A}{\varepsilon_o}$

$E = \frac{\sigma }{2 \varepsilon_o}$

If the charge is negative , $\vec{E}$ will be toward the sheet.

Since in this nature nothing is infinitely large, but this assumption is valid if we placed a unit point charge in front of large sheet.

Applications of Gauss Law(V)

Electric field due to uniform charged sphere-

Sphere may be hollow or solid and both hollow and solid sphere mey be conducting or non-conducting. So let us know the electric field due to all these cases -

In case of conducting sphere, whole charge will come on the surface of the sphere but when the sphere is non-conducting then the whole charge is distributed all over the sphere.

Electric field due to hollow conducting/ Non-conducting and solid conducting sphere -

1. For a point outside the sphere -

$\\ \begin{array}{l}{\text { We first consider the field outside the conductor, so we choose }} {r>R . \text { The entire conductor is within the Gaussian surface, so the }} \\ {\text { enclosed charge is } q \text { . The area of the Gaussian surface is } 4 \pi r^{2} \text { ; }} \\ {\vec{E} \text { is uniform over the surface and perpendicular to it at each point. }} {\text { The flux integral } \oint E_{\perp} d A \text { in Gauss's law is therefore } E\left(4 \pi r^{2}\right)} \\ {\text { which gives }} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\qquad E\left(4 \pi r^{2}\right)=\frac{q}{\varepsilon_{0}}} \\ \\ {\text { or, } \quad E=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}} \quad \text { (outside a charged conducting sphere) }}\end{array}$

For all points outside the shell, field due to uniformly charge shell is such that the entire charge is concentrated at the center of shell.

2. For a point inside the sphere (r<R) -

Since charge will be zero in this case for hollow sphere and conducting solid sphere. So, E = 0 in inside the sphere.

3. At surface (r = R)

$E=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R^{2}}$

Electric field due to solid non-conducting -

1. Outside (r>R)

$E = \frac{\rho R^3}{3 \varepsilon_or^2}$

2. Inisde (r<R) -

$E = \frac{\rho r}{3 \varepsilon_o}$

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Gauss law
Applications of Gauss Law(I)
Applications of Gauss Law(II)

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Books

Reference Books

Applications of Gauss Law(III)

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