Electrostatic Potential and Capacitance Test - 68

Charge will induce on A but total charge on A will remain zero, Negative charge of A will be more closer to B than positive charge on A So potential of B will decrease.

At $$ A  : V_A = k (  \dfrac {q_1}{r_1} + \dfrac {q_2}{r_2})= k( \dfrac { 2.40 \times 10^{-9} C }{ 0.05 m } + \dfrac { -6.50 \times 10^{-9} C} {0.05 m  } ) = -738 V $$

The arrangement shown in the figure is equivalent to the two capacitor joined in parallel. The area of plate in each capacitor is $$A/2$$ :
$$C_1 = \dfrac{\varepsilon_0A}{2d}$$
 $$C_2 = \dfrac{\varepsilon _0\epsilon_rA}{2d}$$
$$C_{eq} = C_1 + C_2 = \dfrac{\varepsilon _0 A}{2d}[1 + \varepsilon_r]$$
$$C = \dfrac{\varepsilon _0 A}{d}$$
$$C_{eq} = \dfrac{C}{2}[1 + \varepsilon_r]$$

Taken without reference to any other point, the potential could have any value.

Because $$C=k\epsilon \,A/d$$ and the dielectric constant $$k$$ increases. So C also increases.

$$Q_A=C_A\ V_A$$
$$=2\times 10^{-6}\times 12$$
$$Q_A=24\times 10^{-6}$$

$$Q_B=C_B\ V_B$$
$$=4\times 10^{-6}\times 6$$
$$Q_B=24\times 10^{-6}$$

Now when plates are attached total charge $$=Q_A+Q_B$$
$$Q=48\times 10^{-6}C$$
Let x he charge on A then Q-x charge is on B
$$ \Rightarrow x=2\mu F\times V\ and\ Q-x=4\mu \ F\times V.$$
$$ \Rightarrow \frac{x}{Q-x}=\frac{1}{2}\ \ \ 2x=Q-x\ \ \ \ x=\frac{9}{3}$$
$$x=16\times 10^{-6};Q-x=32\times 10^{-6}$$
Now charge flows from A to B is given by,
$$=final -initial$$
$$=32\times 10^{-6}-20\times 10^{-6}$$

$$ \bigtriangleup Q=8\times 10^{-6} C$$
$$\bigtriangleup Q = 8\mu C$$

Correct answer: Option C

Hint: A capacitor is defined as an electrical component which stores energy electrostatically in an electric field.

Step 1: Finding final energy

The capacitor originally had the energy U and the work required to pull the insulator out is 4U so the final energy is equal to 5U.

$$W + U = {U_t}$$

$${U_t} = 4U + U$$

$${U_t} = 5U$$

Step 2: Stored energy in terms of capacitance

The capacitance of the insulator with an insulator,

$$C = \dfrac{{A{\varepsilon _0}}}{d}k$$

The formula of the energy stored in the insulator is given by,

$$U = \dfrac{{{q^2}}}{{2C}}$$

Where capacitance is C the area of the insulator is A the dielectric constant is k the distance between the plates is d and k is permittivity of dielectric constant.

When we remove the insulator from the capacitor then the capacitance gets $${C_0}$$

$$U=\dfrac{{{q^2}}}{{2C}}$$

$${C_0} = \dfrac{{A{\varepsilon _0}}}{d}$$

And the stored energy will be,

$${U_t} = \dfrac{{{q^2}}}{{2{C_0}}}$$

Since,

$${U_t} = 5U$$

Step 3: Substituting values

Replacing the value of $$U$$ Ut=5U and $${U_t}$$ Uin the above equation we get,

$$(\dfrac{{{q^2}}}{{2{C_0}}}) = 5(\dfrac{{{q^2}}}{{2C}})$$

$$\dfrac{C}{{{C_0}}} = 5$$

Replacing the value of $$C$$ and $${C_0}$$ in the above relation

$$\dfrac{{\dfrac{{A{\varepsilon _0}}}{d}k}}{{\dfrac{{A{\varepsilon _0}}}{d}}} = 5$$

$$k = 5$$

The dielectric constant value is equal to 5