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Earthing Of Charged Or Uncharged Metal Bodies

Earthing Of Charged Or Uncharged Metal Bodies :- In electrical analysis, the earth is considered as a very large conducting sphere, whose radius is about 6400 kilometers. If the earth is given a charge Q, then its potential becomes :

$\displaystyle {{V}_{E}}=\frac{kQ}{{{R}_{E}}}$

Since the value of is very large, the value of is negligible (very small). So for objects of very small size compared to the earth, we can assume that the potential of the earth is always zero.

If we connect a small object to the earth, the charge flows between the object and the earth until the electric potential of both becomes equal (i.e. zero). Whether the charge flows into the earth or out of the earth, the potential of the earth always remains zero. This means that if a system is at positive potential and is connected to the earth, then the earth will give some negative charge to that system so that ultimately the potential of that system becomes zero.

Consider a neutral solid conducting sphere of radius R with a point charge q placed at a distance x from its centre, as shown in the figure.

Due to induction from the charge q, charges will be induced on the conducting sphere as shown in the above figure. Electric potential on the sphere due to the charge q :

$\displaystyle V=\frac{kq}{x}$

[Since the surface of the conductor is an equipotential surface, the potential at every point on the sphere is the same and hence the distance x from the centre is taken as convenience to find the potential of the sphere due to charge q.]

Here, induced charges have been neglected to find the electric potential of the sphere, because the total potential due to induced charges will be zero.

Electric potential at point A due to induced charge (q_in) :

$\displaystyle {{V}_{A}}=V$

$\displaystyle \Rightarrow \left( \frac{kq}{x-R} \right)+{{\left( {{V}_{A}} \right)}_{in}}=\frac{kq}{x}$

$\displaystyle \Rightarrow {{\left( {{V}_{A}} \right)}_{in}}=kq\left[ \frac{1}{x}-\frac{1}{x-R} \right]$

$\displaystyle \therefore {{\left( {{V}_{A}} \right)}_{in}}=-\left[ \frac{kqR}{x\left( x-R \right)} \right]$

Electric potential at point B due to induced charge (q_in) :

$\displaystyle {{V}_{B}}=V$

$\displaystyle \Rightarrow \left( \frac{kq}{\sqrt{{{x}^{2}}+{{R}^{2}}}} \right)+{{\left( {{V}_{B}} \right)}_{in}}=\frac{kq}{x}$

$\displaystyle \Rightarrow {{\left( {{V}_{B}} \right)}_{in}}=kq\left[ \frac{1}{x}-\frac{1}{\sqrt{{{x}^{2}}+{{R}^{2}}}} \right]$

$\displaystyle \therefore {{\left( {{V}_{B}} \right)}_{in}}=kq\left[ \frac{\sqrt{{{x}^{2}}+{{R}^{2}}}-x}{x\sqrt{{{x}^{2}}+{{R}^{2}}}} \right]$

Electric potential at point C due to induced charge (q_in) :

$\displaystyle {{V}_{C}}=V$

$\displaystyle \Rightarrow \left( \frac{kq}{x+R} \right)+{{\left( {{V}_{C}} \right)}_{in}}=\frac{kq}{x}$

$\displaystyle \Rightarrow {{\left( {{V}_{C}} \right)}_{in}}=kq\left[ \frac{1}{x}-\frac{1}{x+R} \right]$

$\displaystyle \therefore {{\left( {{V}_{C}} \right)}_{in}}=\left[ \frac{kqR}{x\left( x+R \right)} \right]$

Now if the switch S is closed and the sphere is grounded, the potential of the sphere will become zero and for this, some amount of negative charge will flow from the earth to the sphere. Let us assume that charge q_E is transferred from the earth to the sphere.

Since the sphere is grounded, the total electric potential of the sphere is :

$\displaystyle V=\frac{kq}{x}+\frac{k{{q}_{E}}}{R}=0$

$\displaystyle \Rightarrow {{q}_{E}}=-\frac{qR}{x}$

It is clear here that the earth has provided negative charge so that a negative potential is created on the conductor and cancel the initial positive potential due to the charge q.

Note
Always remember that when a metallic body is connected to the earth, the earth either provides or removes a charge (q_E) so that the total electric potential at the body, considering all nearby charges and the body’s own charge, becomes zero.

An example on earthing of charged or uncharged metal bodies to solve numerical problems

As shown in the figure, three concentric thin spherical conducting shells A, B and C are considered with radii a, b and c respectively. Shells A and C are given charges and respectively, and shell B is earthed.

Our objective is to determine the charges on the inner and outer surfaces of the shells A, B and C. To solve such problems, it is important to keep the following points in mind :

(1). The entire charge $q_{A}$ will remain on the outer surface of the shell A, unless some charge is placed inside it. To understand this, let us consider an imaginary spherical Gaussian surface located within the material of the shell A.

Since the electric field inside a conducting material is zero, the total flux passing through this Gaussian surface will also be zero. According to Gauss’s law, the total charge enclosed by this surface should also be zero. Therefore, the entire charge of the shell A comes on its outer surface.

(2). Similarly, if we consider an imaginary spherical Gaussian surface inside the material of the shell B, then :

q₁ + $q_{A}$ = 0

⇒ q₁ = – $q_{A}$ $_{\dots..(1)}$

And similarly, considering an imaginary spherical Gaussian surface in the material of cell C :

q₃ + $q_{2}$ + $q_{1}$ + $q_{A}$ = 0

⇒ q₃ = – $q_{2}$ $_{\dots..(2)}$

(3). Since the charge given to the shell C was , therefore

$q_{3}$ + $q_{4}$ = $_{\dots..(3)}$

(4). Since shell B is earthed, the potential of B must be zero. So

$\displaystyle {{V}_{B}}=k\left[ \frac{{{q}_{A}}}{b}+\frac{({{q}_{1}}+{{q}_{1}})}{b}+\frac{({{q}_{3}}+{{q}_{4}})}{c} \right]=0$ $_{\dots..(4)}$

Using above equations (1), (2), (3) and (4) we can find the values of charges on different surfaces.

In short :

The total charge enclosed within the conductor by a Gaussian surface is zero.
The potential of a grounded conductor is zero.
Contacting conductors are at the same potential.
The charge on all conductors remains constant except those connected to earth.
The charge on the inner surface of the innermost shell will remain zero provided no charge is placed inside it.
Charges of equal magnitude and opposite nature are induced on opposite surfaces.

Example 1.

A charge q is uniformly distributed on the surface of a sphere of radius R. It is covered by a concentric hollow conducting sphere of radius 2R. If the hollow sphere is grounded, then the charge on its outer surface :

(A) q/2

(B) 2q

(D) Zero

Solution :

If we draw an imaginary spherical Gaussian surface within the material of the outer shell, we find that the total charge enclosed by it will be zero only if a charge of -q is induced on the inner surface of the outer shell. Therefore, the inner surface of the outer shell carries a charge of -q.

Let there be a charge q’ on the outer surface of the outer shell. Now since it is grounded, therefore

$\displaystyle V=k\left[ \frac{q}{2R}+\frac{\left( -q \right)}{2R}+\frac{q'}{2R} \right]=0$

$\displaystyle \Rightarrow q'=0$

Therefore there will be no charge on the outer surface of the external shell.

Example 2.

The figure shows three concentric thin spherical conducting shells A, B and C with radii a, b and c respectively. Shells A and C are given charges q and −q respectively and shell B is grounded.

The charge on the outer surface of the shell C will be :

(A) $\displaystyle \frac{b}{c}q$

(B) $\displaystyle \left( 1-\frac{b}{c} \right)q$

(D) $\displaystyle -\left( 1-\frac{b}{c} \right)q$

Solution :

If we draw an imaginary spherical Gaussian surface inside the material of shell B, we find that the total charge enclosed by it will be zero only when a charge of -q is present on the inner surface of the outer shell. Now, if we assume that the outer surface of shell B carries a charge q′, then the inner surface of shell C must have a charge of -q′, and the charge on its outer surface will be [-q – (-q’)] = (q’ – q).

Since shell B is earthed, therefore

$\displaystyle {{V}_{B}}=k\left[ \frac{q}{b}+\frac{\left( -q \right)}{b}+\frac{q'}{b}+\frac{(-q')}{c}+\frac{(q'-q)}{c} \right]=0$

$\displaystyle \Rightarrow \frac{q'}{b}-\frac{q'}{c}+\frac{q'}{c}-\frac{q}{c}=0$

$\displaystyle q'=\frac{b}{c}q$

Hence the charge on the outer surface of shell C :

$\displaystyle q'-q=\frac{b}{c}q-q=\left( \frac{b}{c}-1 \right)q=-\left( 1-\frac{b}{c} \right)q$

Earthing Of Charged Or Uncharged Metal Bodies

An example on earthing of charged or uncharged metal bodies to solve numerical problems

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