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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)

Hence the 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.

Example
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

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

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

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

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

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 :-

Parallel Conducting Plates

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