Electrostatics-Part 2(GAUSS LAW)

Electrostatics-Part 2(GAUSS LAW)

In previous article we have discussed basics of electrostatics. In this article we will cover Gauss law, A very important concept . We will use analytical method to study this , perhaps you would have never seen such an explanation in any coaching classes . This article is contributed by Department of Physics , ABC CLASSES by a combined efforts of entire team.

Gauss Law of Electrostatics

The surface integral of the normal component of Electric field E over any closed surface in an electrostatic field equals times the total or net charge contained in a volume enclosed by the surface.

Where
is the solid angle subtended by area element dS at the position of the charge.

Now,

For a system of discrete charges

Where summation is over only those charges which are inside the surface S.

Flux of Electric field

Flux of electric field E across a surface S is defind as:

where is unit vector in direction of outward drawn normal to the surface.

Two point charges q and –q are separated by the distance 2l (Fig.). Find the flux of the electric field strength vector across a disc of radius R.

Solution :

Let us find flux across disc due to charge q only.

Now, electric field due charge q is not same at every point on disc. So, we will take an element on which E is same.

Let us take ring of radius r and thickness dr

E on any point of the ring.

Total flux due to both charges will be twice due to one charge only.

So,

Application of Gauss Law

(a) To find flux of electric field.

(b) To find electric field in case of symmetric charge distribution.

(c) To find charge distribution for a system of conductors.

Application of Gauss law to calculate Flux of Electric field

(1)

A charge Q is placed at a distance a/2 above the centre of a horizontal, square surface of edge a as shown in figure. Find the flux of the electric field through the square surface.

Solution :

Gauss Law is applicable only in case of closed surface.

Let us make a cube by using eight such surfaces with charge at the centre.

Net flux across the cube will be

Since, q is placed at the centre of symmetry.

So, flux due to q across each faces will be equal.

Hence, flux across each face

(2)A charge q sits at the back corner of a cube, as shown in Fig. What is the flux of E through the shaded side?

Let us add seven more cube of edge a to form a bigger cube of edge 2a such that charge lie at the centre of bigger cube.

So, as we saw in last question, the flux through one of the face is equal to . But we have to find the flux through only ¼ of the face of bigger cube

So, flux through the shaded part

Application of Gauss law to calculate electric field

Before applying Gauss Law for calculation of electric field, Let us discuss about the considerations before using Gauss Law.

(1) Gaussian surface should be symmetric with respect to charge distribution.

(2) Gaussian surface should pass through the point at which electric field is to be found.

Gauss Law in itself does not put these restrictions. But, it is our convenience in solving Gauss Law that we take these considerations.

(3)

Find the electric field due to a non-conducting sheet of charge density

Solution :

Let us assume that non-conducting sheet is positively charged.

Let us take a Gaussian surface in the form of a cylinder with plane bases. Flux across the closed surface

(as is perpendicular to the curved surface)

Now, applying Gauss Law

(4)

Find electric field due to a conducting sheet having surface charge density .

Solution : Discussion

Whenever we give free charge q to the conducting surface, it lies on the surface. There can be no free charge inside conducting volume. Hence, using Gauss Law, electric field inside the conductor is always zero. If we give charge to conducting sheet, however thin it may be, it will always distribute on both the surfaces creating an inner volume between two surfaces where electric field is zero and no free charges. So, we can always visualize a conducting surface to be equivalent to two non-conducting surface.

To evaluate electric field, we can any one surface 1 or 2 as Gaussian surface.

E = 0 inside conductor.

So, if we take 1 as Gaussian surface.

Then ( for curved surface is zero as )

E = 0 at the base inside the conductor

Charge enclosed =

So, using Gauss Law for electrostatics

outside the conductor

We can also obtain taking 2 as Gaussian surface. In that case

Charge enclosed =

So, using Gauss Law for electrostatics.

As we have dealt earlier, a conducting sheet can be visualised as equivalent to two non-conducting sheet. We can use this approach to calculate electric field.

(5)

A non-conducting sphere of radius R is uniformly charged with volume charge density . Find the electric field E inside and outside sphere.

Solution :

To find Electric field E inside the charged sphere, we will take an imaginary surface of radius r < R concentric to the charged sphere.

Let us apply Gauss Law for this surface. Electric field being a central field is constant at the Gaussian surface.

Electric field vector

So,

Using Gauss Law:

× Charge enclosed by Gaussian Surface.

To find electric field outside charged sphere of radius R, we take Gaussian sphere of radius r > R.

Total charge enclosed

Flux enclosed

(6)A ball of radius R carries a positive charge whose volume density depends only on a separaton r from the ball’s centre as , where is a constant. Assuming the relative permittivities of the ball and the environment to be equal to unity, find :

(a)    the magnitude of the electric field strength as a function of the distance r both inside and outside the ball;

(b)the maximum intensity and the corresponding distance .

solution

Volume Charge density

Here varies with r

To find electric field inside the sphere of radius R let us take a Gaussian sphere of radius r

Flux across the Gaussian surface S

Now, Electric field is same at each pointon Gaussian surface and also is parallel to

So, ……..(1)

Now, let us calculate charge enclosed by Gaussian surface s of radius r

Consider a spherical shell of radius x and thickness dx charge contained by spherical shell

So, charge enclosed by the Gaussian sphere of radius r is given by

……..(2)

Applying Gauss Law

……..(3)

Now, Let us find electric field outside the ball let us take Gaussian sphere of radius r (r>R)

Flux P across the Gaussian surface

Change enclosed will be calculated by putting r = R in

equation (2)

Using Gauss Law of electrostatics

So, Electric Field is given by

To calculate main value of electric field

So, Electric field is maximum at r =

(7)A space is filled up with a charge with volume density

, where and are positive constants, r is the distance from the centre of this system. Find the magnitude of the electric field strength vector as a function of r. Investigate the obtained expression for the small and large values of r, i.e. at and .

solution

To find electric field at a distance r form the origin let us take a Gaussian sphere of radius r

Flux across the Gaussian surface

……..(1)

Charge enclosed by Gaussian surface of radius r

Let us solve

Take

So, ……..(2)

Applying Gauss Law

For small value of r, r³<<1

So, for r³<<1

For large value of r

r³>>1

e^-r3<<1

So, E =

(6)

Inside a ball charged uniformly with volume density there is a spherical cavity. The centre of the cavity is displaced with respect to the centre of the ball by a distance a. Find the field strength E inside the cavity, assuming the permittivity equal to unity.

Solution :

Electrostatics is a linear system. So, principle of superposition
can be applied.

The    given    charge    distribution    is    visualised    as    the    superposition
of two charged sphere with charge density and .

Let us find electric field at point inside the cavity.

Electric field E1 due to sphere with charge density

Electric field E2 due to sphere with charge density

Applying principle of superposition

(7)Find the electric field inside cylinder of radius R which is uniformly charged with charge density

Solution

Let us take Gaussian surface in the form of cylinder of radius r (r < a) and length l

Flux enclosed by cylinder

So,

Charge enclosed

Using Gauss Law

Flux enclosed by Gaussian Surface =

In vector notation

(6) Inside an infinitely long circular cylinder charged uniformly with volume density there is a circular cylindrical cavity. The distance between the axes of the cylinder and the cavity is equal to a. Find the electric field strength E inside the cavity. The permittivity is assumed to be equal to unity.

Solution

Fig 1

If we see the top view

Fig.2

The charge configuration in fig (1) can be visualised
as super position of cylinder
with change density and a carry charge with charge density -

Fig.

Electric at Point P will be vector sum of electric field due to cylinder with change density & cavity cylinder with change density -

(7)Two spheres, each of radius R and carrying uniform charge densities and , respectively, are placed so that they partially overlap (Fig.). Call the vector from the positive center to the negative center d. Show that the field in the region of overlap is constant, and find its value.

In this question we will apply a principle of superposition to find electric field at pt P.

Fig.

Electric field at point P will be equal to vector sum of electric field due to positive and negative sphere

Now, at point P

Another application of Gauss Law is to determine the charge distribution especially in case of conductors. Along with the Gauss Law.

We also apply the fact that, electric field inside the conductor is zero.

(8)An uncharged spherical conductor centered at the origin has a cavity of some weird shape carved out of it (Fig.). Somewhere within the cavity is a charge q.

Question: What is the field outside the sphere?

Fig.

Here we will use Gauss Law of electrostatics and fact that electric field inside the conductor is zero.

Since electric field at any point inside the conductor is zero. Hence flux of electric field across any surface enclosing the cavity is zero.

So, it will always enclose, zero change according to Gauss Law of electrostatics.

This means the inner surface will have change induced on it which will be equal to q and opposite in sign.

Now since, conductor is neutral. So, there will be change q on the outer surface.

Now, to find electric field at any point P outside spherical conductor, let us take a Gaussian sphere concentric to spherical conductor and passing though point P.

Flux =

Net change enclosed = q

So, using Gauss Law

(14) A metal sphere of radius R, carrying charge q, is surrounded by a thick concentric metal shell (inner radius a, outer radius b, as in Fig.). The shell carries no net charge.

(a)    Find the surface charge density at R, at a, and at b.

(b)    Find the potential at the center, using infinity as the reference point.

(c)    Now the outer surface is touched to a grounding wire, which lowers its potential to zero (same as at infinity). How do your answers to (a) and (b) change?

There wont be any free change inside the conductor. So, all the charge will be on the surface of conductor so as to ensure zero electric field inside it.

Now, flux across any Gaussian surface which completely we inside the metal shell will be zero because electric field inside metal shell will be zero.

So, this Gaussian surface wont enclose any net change So, change -q will include on the inner surface of metal shell.

Since, shell is neutral, so, outer surface will contain +q charge

Now, electric field

So electric potential at the centre

Now,

Now its outer surface is touched to grounding wire to the potential of the outer surface becomes zero. Here, I would like to discuss the effect of grounding any surface.

Grounding any surface does not necessarily
means that charge on the surface becomes zero. as the common misconception runs.

Grounding just makes the potential of surface zero. If zero potential requires the change on surface to become zero then, charge will become zero if otherwise charge may not be zero.

Ground is unlimited source and unlimited sink of charge.

So, if outer surface of shell is grounded, let the charge on outer surface be q :

Now, q is that value of charge for which potential at the outer surface becomes zero

Now, potential at outer surface =

Since, potential at outer surface = 0

So,    

Hence,            q = 0

(15)

Two conducting plates A and B are placed parallel to each other. A is given a charge Q₁ and B a charge Q₂. Find the distribution of charges on the four surfaces.

Solution : Consider a Gaussian surface as shown in figure (30-W6a). Two faces of this closed surface lie completely inside the conductor where the electric field is zero. The flux through these faces is, therefore, zero. The other parts of the closed surface which are outside the conductor are parallel to the electric field and hence the flux on these parts is also zero. The total flux of the electric field through the closed surface is, therefore, zero. From Gauss’s law, the total charge inside this closed surface should be zero. The charge on the inner surface of A should be equal and opposite to that on the inner surface of B.

The distribution should be like the one shown in figure (30-W6b). To find the value of q, consider the field at a point P inside the plate A. Suppose, the surface area of the plate (one side) is A. Using the equation , the electric field at P

due to the charge (downward),

due to the charge (upward),

due to the charge (downward),

and due to the charge (upward).

The net electric field at P due to all the four charged surfaces is (in the downward direction)

As the point P is inside the conductor, this field should be zero. Hence,

or,         … (i)

Thus,         … (ii)

and         … (iii)

Using these equations, the distribution shown in the figure (30-W6) can be redrawn as in figure (30-W7).

This result is a special case of the following result. When charged conducting plates are placed parallel to each other, the two outermost surfaces get equal charges and the facing surfaces get equal and opposite charges.

(15)

Two large conducting plates are placed parallel to each other and they carry equal and opposite charges with surface density as shown in figure .Find the electric field

(a)at the left of the plates

(b) in between the plates and

(c) at the right of plates

Solution

At point P at the left of plates

Electric field will be super position of electric field due to positive change and negative charge

Similary at pt Q

At R.

(16)

Two conducting plates X and Y each having large surface area A (on one side ),are placed parallel to each other as shown in figure. The plate X is given a charge Q whereas the other is neutral. Find (a) the surface charge density at the inner surface of the plateX

(b)The electric field at a point to the left of the plates

(c) The electric field at a point in between the plates and

(d) the electric field at a point to the right of the plates

Solution

Let the charge on inner face of X is q then charge on left face X is Q–q

Using Gauss Law as in last question

charge on left face of Y is –q

Since, Y is neutral, change on right face of Y is q

Now we need to determine q

The electric field inside the any conductor is zero. We will exploit this fact to find the value of q.

Electric field at print P will be superposition of all the four faces shown in figure.

So,

So, surface change density at the inner surface of the plate X =

Assume Q to be positive

Electric field at a point left of the plates

toward left

Fig.

Electric field at any point between the plates

towards right

Electric field at a point to the right of the plates

towards right

(24)

Three identical metal plates with large surface areas are kept parallel to each other as shown in fig.The leftmost plate is given a charge Q,the rightmost a charge -2Q and the middle one remains neutral .Find the charge appearing on the outer surface of the rightmost plate

Fig.

Solution

Here also, we have used Gauss Law to distribute charge on the plate.

Now we will make use of the fact that electric field inside metal is zero.

Electric field at point P is superposition of electric field due to all the six surfaces



Change appearing on the outer surface of the right most plate

We will discuss Work & Energy in Next Article

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