Mix Examples - Electric Potential and Capacitance Questions in English

(A) Let the initial separation be $r_i = 4R$ and the final separation when they just touch be $r_f = 2R$.
By the principle of conservation of energy,the total initial energy equals the total final energy.
Initial Energy: $E_i = 2 \times (\frac{1}{2} m u^2) - \frac{G m^2}{4R} + \frac{k Q^2}{4R} = m u^2 - \frac{G m^2}{4R} + \frac{k Q^2}{4R}$.
Final Energy (at the moment of touching,speed is zero): $E_f = 0 - \frac{G m^2}{2R} + \frac{k Q^2}{2R}$.
Equating $E_i = E_f$:
$m u^2 - \frac{G m^2}{4R} + \frac{k Q^2}{4R} = - \frac{G m^2}{2R} + \frac{k Q^2}{2R}$.
$m u^2 = \frac{k Q^2}{2R} - \frac{k Q^2}{4R} - \frac{G m^2}{2R} + \frac{G m^2}{4R}$.
$m u^2 = \frac{k Q^2}{4R} - \frac{G m^2}{4R} = \frac{1}{4R} (k Q^2 - G m^2)$.
$u^2 = \frac{1}{4 m R} (k Q^2 - G m^2) = \frac{k Q^2}{4 m R} (1 - \frac{G m^2}{k Q^2})$.
$u = \sqrt{\frac{k Q^2}{4 m R} (1 - \frac{G m^2}{k Q^2})}$.

(A) The electrostatic energy $U$ of a capacitor connected to a battery is $U = \frac{1}{2}CV^2$.
Here $C = \frac{\epsilon_0 A}{x}$,so $U = \frac{1}{2} \left( \frac{\epsilon_0 A}{x} \right) V^2$.
Since the battery is connected,the potential difference $V$ remains constant.
Therefore,$U \propto \frac{1}{x} = x^{-1}$.
The time rate of change of energy is $\frac{dU}{dt} = \frac{dU}{dx} \cdot \frac{dx}{dt}$.
Given that the plates are pulled apart at a uniform speed $v$,we have $\frac{dx}{dt} = v$ (a constant).
Now,differentiating $U$ with respect to $x$: $\frac{dU}{dx} = \frac{d}{dx} \left( \frac{\epsilon_0 A V^2}{2} \cdot x^{-1} \right) = -\frac{\epsilon_0 A V^2}{2} \cdot x^{-2}$.
Thus,$\frac{dU}{dt} = \left( -\frac{\epsilon_0 A V^2}{2} \cdot x^{-2} \right) \cdot v$.
Since $\epsilon_0, A, V,$ and $v$ are constants,we get $\frac{dU}{dt} \propto x^{-2}$.
Comparing this with $x^\alpha$,we find $\alpha = -2$.

(B) Assertion $A$ is true: According to Gauss's law,the electric field inside a conductor in electrostatic equilibrium is zero,which implies that the net charge density inside the conductor is zero.
Reason $R$ is also true: If a free charge carrier is placed between the plates of a capacitor (in a vacuum or air),it experiences an electric force $F = qE$ and undergoes drift.
However,the reason for the absence of net charge inside a conductor is the redistribution of charges on the surface to cancel the internal field,which is independent of the behavior of charges between capacitor plates. Therefore,$R$ is not the correct explanation of $A$.

(D) The circuit is a balanced Wheatstone bridge because $\frac{C_1}{C_2} = \frac{C_4}{C_3} = \frac{10}{10} = 1$.
Since the bridge is balanced,no charge flows through the central capacitor $C_5$.
Therefore,the circuit simplifies to two parallel branches,each containing two capacitors in series.
The left branch consists of $C_1$ and $C_2$ in series,and the right branch consists of $C_4$ and $C_3$ in series.
Equivalent capacitance of each branch: $C_{\text{branch}} = \frac{10 \times 10}{10 + 10} = 5 \mu F$.
Total equivalent capacitance: $C_{\text{eq}} = 5 \mu F + 5 \mu F = 10 \mu F$.
Total charge supplied by the battery: $Q = C_{\text{eq}}V = 10 \mu F \times 50 \ V = 500 \mu C$.
Since the branches are identical,the charge divides equally: $250 \mu C$ flows through each branch.
Thus,$250 \mu C$ is on each capacitor $C_1, C_2, C_3, C_4$ and $0 \mu C$ is on $C_5$.