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Parallel Conducting Plates

Parallel Conducting Plates :- Let us consider two conducting plates parallel to each other. Plate I is given a charge Q₁ and plate II is given a charge Q₂ which distributes itself as shown in figure below :-

here

q₁ + q₂ = Q₁ …..(1)

q₃ + q₄ = Q₂ …..(2)

Now consider a Gaussian surface ABCD as shown in figure. Two faces of this closed surface lie completely inside the conductor(AD and BC) 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 ABCD is, therefore zero. From Gauss’s law, the total charge inside this closed surface should be zero. The charge on the inner surface of I should be equal and opposite to that on the inner surface of II.

Therefore,

q₂ = – q₃ …..(3)

Now to find further relation between the distributed charges we find electric field at point P.

We know that electric field due to a single layer of charge is $\displaystyle \frac{\sigma }{2{{\varepsilon }_{0}}}=\frac{q}{2A{{\varepsilon }_{0}}}$

Electric field at point P :-

Due to q₁ layer of charge = $\displaystyle \frac{{{q}_{1}}}{2A{{\varepsilon }_{0}}}$ (towards right hand side)

Due to q₂ layer of charge = $\displaystyle \frac{{{q}_{2}}}{2A{{\varepsilon }_{0}}}$ (towards left hand side)

Due to q₃ layer of charge = $\displaystyle \frac{{{q}_{3}}}{2A{{\varepsilon }_{0}}}$ (towards left hand side)

Due to q₄ layer of charge = $\displaystyle \frac{{{q}_{4}}}{2A{{\varepsilon }_{0}}}$ (towards left hand side)

Net electric filed at point P,

$\displaystyle {{\left( {{E}_{net}} \right)}_{at\text{ }P}}=\frac{{{q}_{1}}}{2A{{\varepsilon }_{0}}}-\frac{{{q}_{2}}}{2A{{\varepsilon }_{0}}}-\frac{{{q}_{3}}}{2A{{\varepsilon }_{0}}}-\frac{{{q}_{4}}}{2A{{\varepsilon }_{0}}}$

As the point P lies inside the conductor, this electric field should be zero, hence

$\displaystyle \frac{{{q}_{1}}}{2A{{\varepsilon }_{0}}}-\frac{{{q}_{2}}}{2A{{\varepsilon }_{0}}}-\frac{{{q}_{3}}}{2A{{\varepsilon }_{0}}}-\frac{{{q}_{4}}}{2A{{\varepsilon }_{0}}}=0$

$\displaystyle \Rightarrow \frac{{{q}_{1}}}{2A{{\varepsilon }_{0}}}=\frac{{{q}_{2}}+{{q}_{3}}+{{q}_{4}}}{2A{{\varepsilon }_{0}}}$

$\displaystyle \Rightarrow {{q}_{1}}={{q}_{2}}+{{q}_{3}}+{{q}_{4}}$

Now as q₂ = – q₃, hence

q₁ = q₄ …..(4)

Now adding equations (1) & (2), we get

$\displaystyle {{q}_{1}}+{{q}_{2}}+{{q}_{3}}+{{q}_{4}}={{Q}_{1}}+{{Q}_{2}}$

$\displaystyle \Rightarrow {{q}_{4}}+(-{{q}_{3}})+{{q}_{3}}+{{q}_{4}}={{Q}_{1}}+{{Q}_{2}}$

$\displaystyle {{q}_{4}}=\frac{{{Q}_{1}}+{{Q}_{2}}}{2}$

Putting the value of q₄ in equations (1) & (2), we get

$\displaystyle {{q}_{2}}=\frac{{{Q}_{1}}-{{Q}_{2}}}{2}$ & $\displaystyle {{q}_{3}}=\frac{{{Q}_{2}}-{{Q}_{1}}}{2}$

Finally,

$\displaystyle {{q}_{1}}={{q}_{4}}=\frac{{{Q}_{1}}+{{Q}_{2}}}{2}$ …..(5)

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

$\displaystyle {{q}_{3}}=\frac{{{Q}_{2}}-{{Q}_{1}}}{2}$ …..(7)

Similarly, let us discuss a case of three parallel conducting plates I, II and III having charges Q₁, Q₂and Q₃ respectively. These charges will be distributed as shown in figure below :

Here

q₁ + q₂ = Q₁ …..(8)

q₃ + q₄ = Q₂ …..(9)

q₅ + q₆ = Q₃ …..(10)

From the above discussion of two parallel conducting plates, for the Gaussian surfaces ABCD and EFGH we can write :

q₂ = – q₃ …..(11)

q₄ = – q₅ …..(12)

Now electric field at point P :-

Due to q₁ layer of charge = $\displaystyle \frac{{{q}_{1}}}{2A{{\varepsilon }_{0}}}$ (towards right hand side)

Due to q₂ layer of charge = $\displaystyle \frac{{{q}_{2}}}{2A{{\varepsilon }_{0}}}$ (towards left hand side)

Due to q₃ layer of charge = $\displaystyle \frac{{{q}_{3}}}{2A{{\varepsilon }_{0}}}$ (towards left hand side)

Due to q₄ layer of charge = $\displaystyle \frac{{{q}_{4}}}{2A{{\varepsilon }_{0}}}$ (towards left hand side)

Due to q₅ layer of charge = $\displaystyle \frac{{{q}_{5}}}{2A{{\varepsilon }_{0}}}$ (towards left hand side)

Due to q₆ layer of charge = $\displaystyle \frac{{{q}_{6}}}{2A{{\varepsilon }_{0}}}$ (towards left hand side)

Net electric filed at point P,

As the point P lies inside the conductor, this electric field should be zero, hence

$\displaystyle \frac{{{q}_{1}}}{2A{{\varepsilon }_{0}}}-\frac{{{q}_{2}}}{2A{{\varepsilon }_{0}}}-\frac{{{q}_{3}}}{2A{{\varepsilon }_{0}}}-\frac{{{q}_{4}}}{2A{{\varepsilon }_{0}}}-\frac{{{q}_{5}}}{2A{{\varepsilon }_{0}}}-\frac{{{q}_{6}}}{2A{{\varepsilon }_{0}}}=0$

$\displaystyle \Rightarrow {{q}_{1}}={{q}_{2}}+{{q}_{3}}+{{q}_{4}}+{{q}_{5}}+{{q}_{6}}$

Using equations (11) and (12) in above equation, we get

$\displaystyle {{q}_{1}}={{q}_{6}}$ …..(13)

Adding equations (8), (9) and (10),

$\displaystyle {{q}_{1}}+{{q}_{2}}+{{q}_{3}}+{{q}_{4}}+{{q}_{5}}+{{q}_{6}}={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}$

Again using equations (11) and (12) in above equation,

$\displaystyle {{q}_{1}}+{{q}_{6}}={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}$ …..(14)

Finally from equations (13) and (14), we can write

$\displaystyle {{q}_{1}}={{q}_{6}}=\frac{{{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}}{2}$ …..(15)

Now using equation (8)

$\displaystyle {{q}_{2}}={{Q}_{1}}-{{q}_{1}}={{Q}_{1}}-\frac{{{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}}{2}=\frac{{{Q}_{1}}-{{Q}_{2}}-{{Q}_{3}}}{2}$ …..(16)

From equation (11)

$\displaystyle {{q}_{3}}=-{{q}_{2}}=\frac{{{Q}_{2}}+{{Q}_{3}}-{{Q}_{1}}}{2}$ …..(17)

Using equation (9)

$\displaystyle {{q}_{4}}={{Q}_{2}}-{{q}_{3}}={{Q}_{2}}-\frac{{{Q}_{2}}+{{Q}_{3}}-{{Q}_{1}}}{2}=\frac{{{Q}_{2}}+{{Q}_{1}}-{{Q}_{3}}}{2}$ …..(18)

And from equation (12)

$\displaystyle {{q}_{5}}=-{{q}_{4}}=\frac{{{Q}_{3}}-{{Q}_{1}}-{{Q}_{2}}}{2}$ …..(19)

From the above discussion, we observe from equations (5) and (15) that the charges on the outermost surfaces of the parallel plate system are equal, and each is equal to half of the total charge of the system.

Hence the result, When charged conducting plates are placed parallel to each other, the two outermost surfaces acquire equal charges, each equal to half of the total charge of the system, while the facing surfaces acquire equal and opposite charges.

Example 1.
Figure shows three large metallic plates with charges – Q, 3Q and Q respectively. Determine the final charges on all the surfaces.

Solution :-

We assume that charge on surface 2 is x. Following conservation of charge, we see that surfaces 1 has charge (– Q – x). The electric field inside the metal plate is zero so fields at P is zero.

Resultant electric field at P is zero, hence

$\displaystyle {{\left( {{E}_{net}} \right)}_{at\text{ }P}}=-\frac{\left( Q+x \right)}{2A{{\varepsilon }_{0}}}-\frac{x}{2A{{\varepsilon }_{0}}}-\frac{3Q}{2A{{\varepsilon }_{0}}}-\frac{Q}{2A{{\varepsilon }_{0}}}=0$

$\displaystyle \Rightarrow -\frac{\left( Q+x \right)}{2A{{\varepsilon }_{0}}}=\frac{x}{2A{{\varepsilon }_{0}}}+\frac{3Q}{2A{{\varepsilon }_{0}}}+\frac{Q}{2A{{\varepsilon }_{0}}}$

$\displaystyle \Rightarrow -\left( Q+x \right)=x+3Q+Q$

$\displaystyle \Rightarrow x=\frac{-5Q}{2}$

[Note :- Here you may think that to find $\displaystyle {{\left( {{E}_{net}} \right)}_{at\text{ }P}}$ we have used the total charge “3Q” and “Q” instead of assuming the charge distribution on the second and third plates. For this you can see Example 4.]

Now we know that charges on the facing surfaces of the plates are of equal magnitude and opposite sign.
Thus the final charge distribution on all the surfaces is :-

Second Method :-

Value of charge on outermost surfaces :

$\displaystyle {{q}_{1}}={{q}_{6}}=\frac{-Q+3Q+Q}{2}=+\frac{3Q}{2}$

Total Charge on 1^st plate is “-Q” so charge on its inner surface :

$\displaystyle {{q}_{2}}=-Q-\frac{3Q}{2}=-\frac{5Q}{2}$

Similarly charges on other surfaces can also be calculated.

Example 2.

Two long parallel conductor plates (both are at a finite distance from each other) are given charges Q and 2Q respectively. Then calculate the charge present on each surface.

Solution :-

The value of charge on the outermost surfaces :

$\displaystyle {{q}_{1}}={{q}_{4}}=\frac{Q+2Q}{2}=+\frac{3Q}{2}$

The total charge on the first plate is “Q”, so the charge on its inner surface :

$\displaystyle {{q}_{2}}=Q-\frac{3Q}{2}=-\frac{Q}{2}$

On opposite surfaces, there is equal and opposite charge, therefore

$\displaystyle {{q}_{3}}=-{{q}_{2}}=+\frac{Q}{2}$

Hence the charge distribution will be as follows :

Example 3.

An isolated conductor plate of charge Q and area A is placed in a uniform electric field E in such a way that the electric field is perpendicular to the plate and is present on the entire plate. Then find the value of the charges generated on both its surfaces.

Solution :-

Suppose the charge on the left face of the plate is x then the charge on the right face will be (Q-x).

Since net electric field in a conductor is zero, therefore at point P in the plate :

$\displaystyle {{\left( {{E}_{net}} \right)}_{at\text{ }P}}=\frac{x}{2A{{\varepsilon }_{0}}}+E-\frac{Q-x}{2A{{\varepsilon }_{0}}}=0$

$\displaystyle \Rightarrow \frac{x}{2A{{\varepsilon }_{0}}}+\frac{x}{2A{{\varepsilon }_{0}}}-\frac{Q}{2A{{\varepsilon }_{0}}}+E=0$

$\displaystyle \therefore x=\frac{Q}{2}-EA{{\varepsilon }_{0}}$

And charge on the second surface,

$\displaystyle Q-x=Q-\left( \frac{Q}{2}-EA{{\varepsilon }_{0}} \right)=\frac{Q}{2}+EA{{\varepsilon }_{0}}$

Hence the charge on one surface is $\displaystyle \frac{Q}{2}-EA{{\varepsilon }_{0}}$ and the charge on the other surface is $\displaystyle \frac{Q}{2}+EA{{\varepsilon }_{0}}$ .

Example 4.

If an isolated infinite length plate has charge Q₁ on one surface and charge Q₂ on the other surface, then prove that the electric field intensity at any point in front of the plate is $\displaystyle \frac{Q}{2A{{\varepsilon }_{0}}}$ . Where Q = Q₁ + Q₂.

Solution :-

Net electric field intensity at any point P near the plate is :

$\displaystyle {{\left( {{E}_{net}} \right)}_{at\text{ }P}}=\frac{{{Q}_{1}}}{2A{{\varepsilon }_{0}}}+\frac{{{Q}_{2}}}{2A{{\varepsilon }_{0}}}=\frac{1}{2A{{\varepsilon }_{0}}}({{Q}_{1}}+{{Q}_{2}})=\frac{Q}{2A{{\varepsilon }_{0}}}$

[This proves that the intensity of the electric field due to the plate depends on the total charge on the plate. It does not depend on the charge distribution on different surfaces.]

Parallel Conducting Plates

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