Question

Three large conducting parallel plates each having same area are connected to a battery. $E_1$ is the energy stored between the plates initially and $E_2$ finally and $F_1$ is the force between plates 1 and 3 initially and $F_2$ is the force between them finally.
Match List $I$ with List $II$ and select the correct answer using the codes given below the lists.

	List $I$	List $II$
p.	Now, the middle plate shifts towards the third plate very slowly keeping $1^{st}$and $3^{rd}$ plates fixed. Initial and final situations shown in figure. Then	1.$\frac{E_1}{E_2}>1$
q.	Three large conducting parallel plates each having same area are connected to the battery and the space between the $2^{nd}$ and $3^{rd}$ plates is filled with dielectric. Now, the dielectric is removed very slowly. Then	2.$\frac{E_1}{E_2}<1$
r.	Three large conducting parallel plates each having same area and having charges $-Q$, $2Q$ and $-Q$ respectively initially. Now first plate shifts towards the left and $3^{rd}$ towards right keeping $2^{nd}$ fixed. Then	3.$\frac{F_1}{F_2}<1$
s.	Three large conducting parallel plates each having same area, charge on $1^{st}$ plate is $-$$Q$, on $2^{nd}$ plate is $Q$ and the third is uncharged. Now, the switch is closed. If $F_1$ is the force between plate 1 and 3 initially and $F_2$ is the force between them finally, then	4.$\frac{F_1}{F_2}=1$

• A. p - 2, q - 1, r - 3, s - 4
• B. p - 1, q - 2, r - 3, s - 4
• C. p - 2, q - 1, r - 4, s - 3
• D. p - 1, q - 2, r - 4, s - 3

Answer 1

The correct option is C p - 2, q - 1, r - 4, s - 3

(a)
$C=\frac{A{ϵ}_0}{d}$
$E_1=\frac{1}{2}c{ϵ}^2+\frac{1}{2}c{ϵ}^2$
$E_2=\frac{1}{2}\frac{2c}{3}{ϵ}^2+\frac{1}{2}2c{ϵ}^2$
Force between plates is given by $F=\frac{Q_1{Q}_2}{2A{ϵ}_0}$

$F_1=\frac{c^2{ϵ}^2}{2A{ϵ}_0}$
$F_2=\frac{\frac{2cϵ}{3}.2cϵ}{2A{ϵ}_0}=\frac{2c^2{ϵ}^2}{3A{ϵ}_0}$

Net charge flowing through the battery $=\frac{2cϵ}{3}$
Work done by battery = $\frac{2cϵ}{3}ϵ=\frac{2}{3}c{ϵ}^2$

(b)

$C=\frac{A{ϵ}_0}{d}$
$E_1=\frac{1}{2}c{ϵ}^2+\frac{1}{2}kc{ϵ}^2$
$E_2=\frac{1}{2}c{ϵ}^2+\frac{1}{2}c{ϵ}^2$
$F_1=\frac{k{c}^2{ϵ}^2}{2A{ϵ}_0}$
$F_2=\frac{c^2{ϵ}^2}{2A{ϵ}_0}$

Net charge flowing through the battery $=-(k-1)cϵ$
Work done by battery is -ve

(c)

Charge doesn't change as it is an isolated system.
$E_1=\frac{Q^2}{2C}+\frac{Q^2}{2C}$
$E_2=\frac{Q^2}{2C_1}+\frac{Q^2}{2C_1}$ ($C_1<C$)
$F_1=\frac{Q^2}{2A{ϵ}_0}$
$F_2=\frac{Q^2}{2A{ϵ}_0}$

(d)

$E_1=\frac{Q^2}{2C}$
$E_2=\frac{Q^2}{8C}+\frac{Q^2}{8C}=\frac{Q^2}{4C}$
$F_1=0$
$F_2=\frac{Q^2/4}{2A{ϵ}_0}=\frac{Q^2}{8A{ϵ}_0}$