Question

Two large, parallel conducting plates X and Y, kept close to each other, are given charges $Q_1$ and $Q_2(Q_1>Q_2)$. The four surfaces of the plates are A, B, C and D, as shown in figure. Then :

• A. The charge on B is $\frac{1}{2}(Q_1-Q_2)$.
• B. The charge on A is $\frac{1}{2}(Q_1+Q_2)$.
• C. The charge on C is -$\frac{1}{2}(Q_1-Q_2)$.
• D. The charge on D is $\frac{1}{2}(Q_1+Q_2)$

Answer 1

Answer: (A), (B), (C), (D)

Let charge for surface A $=Q_1-q$, for surface B $=q$, for surface C $=Q_2-q^{\prime }$ and for surface D $=q^{\prime }$.
As the field inside X plate is zero, $\frac{1}{2A{ϵ}_0}(Q_1-q-q-Q_2+q^{\prime }-q^{\prime })=0$
$-2q=Q_2-Q_1\Rightarrow q=\frac{1}{2}(Q_1-Q_2)$
similarly for Y plate, $\frac{1}{2A{ϵ}_0}(q^{\prime }-Q_2+q^{\prime }-q-Q_1+q)=0$
or $2q^{\prime }=Q_1+Q_2\Rightarrow q^{\prime }=\frac{1}{2}(Q_1+Q_2)$
Thus, charge on A is $Q_1-q=Q_1-\frac{1}{2}(Q_1-Q_2)=\frac{1}{2}(Q_1+Q_2)$.
charge on B is $q=\frac{1}{2}(Q_1-Q_2)$
charge on C is $Q_2-q^{\prime }=Q_2-\frac{1}{2}(Q_1+Q_2)=-\frac{1}{2}(Q_1-Q_2)$
Charge on D is $q^{\prime }=\frac{1}{2}(Q_1+Q_2)$