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Earthing of Parallel Plates | Earthing of Parallel Conducting Plates

Earthing of Parallel Plates | Earthing of Parallel Conducting Plates :- Before discussing “Earthing of Parallel Plates” , let us first discuss single earthed metal plate. Let us assume that a metal plate is given a charge +Q. This charge +Q is equally divided on both of the surfaces of the plate as shown in figure below.

If this plate is now earthed, then its potential will become zero(as the potential of the Earth is zero).

For the potential to become zero, the whole charge +Q of the plate will flow to earth(in other words electrons will flow from the earthing wire to the plate, i.e., -Q charge is supplied by the Earth to the plate).

Now suppose a system of two thin parallel metal plates out of which one plate A is given a charge +Q which will be divided equally on both of the surfaces of the plate. Now due to induction, charges +Q/2 and -Q/2 will induce on both of the surfaces of the other plate B, as shown in figure below.

Here the potential of plate B is positive, which is due to +Q charge of plate A (note that potential of plate B is zero due to induced charges +Q/2 and -Q/2).

Now if the plate B is earthed by switch S, then its potential will become zero. First the charge +Q/2 of plate B will flow to the Earth(actually earth supplies a charge of -Q/2 to nullify the charge +Q/2) and charge -Q/2 of plate B remains on it as it is bound due to +Q/2 charge of plate A.

Now the potential of plate B is negative due to charge -Q/2 on it and positive due to charge +Q of plate A. So an extra -Q/2 charge is required for the potential of plate B to become zero. As some extra negative charge starts collecting on plate B from the Earth, the overall negative charge of plate B starts increasing and this extra negative charge starts pulling the +Q/2 charge on left side of plate A to the right side of plate A. Finally when extra -Q/2 charge has collected on plate B from the Earth, all of +Q/2 charge of plate A is pulled on its right side.

At the end the charge distribution on plates is as shown below :-

Electric field between the plates before earthing,

$\displaystyle {{E}_{initial}}=\frac{\left( Q/2 \right)}{2A{{\varepsilon }_{0}}}+\frac{\left( Q/2 \right)}{2A{{\varepsilon }_{0}}}=\frac{Q}{2A{{\varepsilon }_{0}}}$ …..(1)

Potential difference between the plates before earthing,

$\displaystyle \Delta {{V}_{\text{intital}}}={{V}_{A}}-{{V}_{B}}={{E}_{initial}}\times d=\frac{Q}{2A{{\varepsilon }_{0}}}d$ …..(2)

Electric filed between the plates after earthing,

$\displaystyle {{E}_{final}}=\frac{Q}{2A{{\varepsilon }_{0}}}+\frac{Q}{2A{{\varepsilon }_{0}}}=\frac{Q}{A{{\varepsilon }_{0}}}=2{{E}_{initial}}$ …..(3)

Potential difference between the plates after earthing,

$\displaystyle \Delta {{V}_{final}}={{V}_{A}}-{{V}_{B}}={{E}_{final}}\times d=\frac{Q}{A{{\varepsilon }_{0}}}d$ …..(4)

As the plate B is earthed, the potential difference between the plates is equal to the potential of plate A, hence

$\displaystyle {{V}_{A}}=\frac{Q}{A{{\varepsilon }_{0}}}d$

Hence just after the earthing, the electric field and potential difference between the plates is doubled.

Now let us consider another example of three parallel thin metal plates A,B and C, out of which middle plate B is given a charge Q. Due to electrostatic induction, charges +Q/2 and -Q/2 are induced on the plates A and C as shown in figure below :-

Now as discussed above if the switch S₁ is closed, whole charge of plate B will shift on its left surface and a charge -Q flows through the switch S₁ from the Earth to plate A and the final situation will be as shown below :-

If instead of switch S₁, switch S₂ is closed, the distribution of charges on the plates will be obviously as shown below :-

If we close both the switches simultaneously, the situation will be according to the figure shown below :-

The charge Q on plate B is distributed as shown in figure above and equal and opposite charges -q₁ and -q₂ are induced on the inner surfaces of plates A and C.

Let us calculate the values of charges -q₁ and -q₂as discussed below :-

$\displaystyle \Rightarrow {{V}_{B}}-{{V}_{A}}={{V}_{B}}-{{V}_{C}}$

This is because both the plates A and C are earthed, hence their potentials are zero.

$\displaystyle \Delta {{V}_{BA}}=\Delta {{V}_{BC}}$

$\displaystyle \Rightarrow {{E}_{1}}{{d}_{1}}={{E}_{2}}{{d}_{2}}$

$\displaystyle \frac{{{q}_{1}}}{A{{\varepsilon }_{0}}}{{d}_{1}}=\frac{{{q}_{2}}}{A{{\varepsilon }_{0}}}{{d}_{2}}$

$\displaystyle \Rightarrow {{q}_{1}}{{d}_{1}}={{q}_{2}}{{d}_{2}}$

$\displaystyle \Rightarrow {{q}_{1}}=\frac{{{q}_{2}}{{d}_{2}}}{{{d}_{1}}}$ …..(5)

And also,

$\displaystyle {{q}_{1}}+{{q}_{2}}=Q$ …..(6)

Solving equations (5) and (6), we get

$\displaystyle {{q}_{1}}=\left( \frac{{{d}_{2}}}{{{d}_{1}}+{{d}_{2}}} \right)Q$ …..(7)

$\displaystyle {{q}_{2}}=\left( \frac{{{d}_{1}}}{{{d}_{1}}+{{d}_{2}}} \right)Q$ …..(8)

Thus if both the switches S₁ and S₂ are closed simultaneously, charges q₁ and q₂ as given by equations (7) and (8) flows from the Earth through the switches S₁ and S₂ to the plates A and C respectively.

Earthing of Parallel Plates | Earthing of Parallel Conducting Plates

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