Question

X and Y are large,parallel conducting plates close to each other. Each face has an area A. X is given a charge Q. Y is without any charge. Points A, B and C are as shown in the figure then:

• A. the field at B is $\frac{Q}{2{\epsilon }_{0}A}$
• B. the field at B is $\frac{Q}{{\epsilon }_{0}A}$
• C. the field at A, B and C are of the same magnitude
• D. the fields at A and C are of the same magnitude, but in opposite directions

Answer 1

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

The correct options are
A the field at B is $\frac{Q}{2{\epsilon }_{0}A}$
C the field at A, B and C are of the same magnitude
D the fields at A and C are of the same magnitude, but in opposite directions

Charge on plate X is $Q_1=Q$ and charge on plate Y is $Q_2=0$

Charges on plate surfaces 1 and 4 are $q_1=q_4=\frac{Q_1+Q_2}{2}=\frac{Q}{2}$

Charge on surface 2 is $q_2=\frac{Q_1-Q_2}{2}=\frac{Q}{2}$

and charge on surface 3 is $q_3=-q_2=-\frac{Q}{2}$

The field at A is $E_A=E_1+E_4=\frac{\left(Q/2\right)}{2A{ϵ}_0}+\frac{\left(Q/2\right)}{2A{ϵ}_0}=\frac{Q}{2A{ϵ}_0}$

As both 1 and 4 surfaces have same positive charge so fields at A and C are directed away from the surfaces and have equal magnitude.so are equal in magnitude but opposite in direction.

Thus, $E_A=\frac{Q}{2A{ϵ}_0}$ (toward left) and $E_C=\frac{Q}{2A{ϵ}_0}$ (toward right)

Now $E_B=E_2+E_3=\frac{Q/2}{2A{ϵ}_0}+\frac{Q/2}{2A{ϵ}_0}=\frac{Q}{2A{ϵ}_0}$ (toward right) as filled is directed from positive to negative charge.