Electric potential and capacitance practice · Medium

Option B (The electric potential is zero) is correct because the electric potential is a scalar quantity and at any point on the right bisector, the potentials due to +q and -q cancel each other.

Option D is correct. By using the given conditions, we can solve for the individual capacitances.

Option D (Information is not enough to give answer) is correct because the potential at a corner and the center of a uniformly charged cube depends on the charge distribution, which is not fully defined.

Option C (Another capacitor should be added in series with the first capacitor) is correct because adding a capacitor in series decreases the equivalent capacitance, thus increasing the resonant frequency.

The question is flawed. The correct answer should be 1/2. The provided options are incorrect and nonsensical. The ratio of the energy stored in the capacitor to the work done by the battery is always 1/2.

Option C (Infinite) is correct because a perfect conductor allows free flow of charge, effectively neutralizing any applied electric field, which corresponds to an infinite dielectric constant.

Option C ($\frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{R}$) is correct because the electric potential inside a hollow conducting sphere is constant and equal to the potential on its surface.

Option D ($R$) is correct because the capacitance of an isolated conducting sphere is directly proportional to its radius.

Option A (Their capacities are equal) is correct because the condition $C_p = 4C_s$ leads to $C_1 = C_2$.

Option D ($\frac{{V}}{{4}}$) is correct because the magnetic potential is directly proportional to the magnetic dipole moment.

Option B ($- 3 \times {10^{ - 23}}J$) is correct because the electrical energy in equilibrium is given by $U = -pE$, where $p = qd$.

Option C ($\frac{6}{{14}}$ and $30volt$) is correct because the charge redistributes proportionally to the capacitance, and the final potential is determined by charge conservation.

Option B is correct. The electric field between the plates decreases by a factor equal to the dielectric constant when the capacitor is immersed in a dielectric.

Option A (If both Assertion and Reason are correct and reason is the correct explanation of Assertion) is correct because the random orientation of dipoles in the absence of an external field results in a net zero dipole moment per unit volume.

Option D ($10/9\;Volt/\mu m$ and in the $+ ve\mathrm{ }\;x$ direction) is correct because it is the result of calculating the negative derivative of the potential function at x=4.

Option D (The potential difference in the free plates becomes $2V$) is correct because when connected in series with opposite polarity, the voltages add up.

Option C is correct because the electric potential due to a point charge is inversely proportional to the distance from the charge, given by $V = \frac{{kQ}}{{r}}$.

Option D ($\frac{1}{2} \mathrm{QV}$) is correct because it represents the standard formula for the energy stored in a capacitor.

Option A (Increases) is correct because the electric potential energy of a positive charge increases when it moves from a lower potential to a higher potential.

Option D (4) is correct because the dielectric constant is calculated by considering the changes in capacitance due to the doubled distance and the liquid immersion.

Option B (Inversely proportional to ${r^3}$) is correct because the electric field due to a dipole decreases with the cube of the distance on its axis.

Option B (Half the energy supplied is stored in the capacitor) is correct because half of the energy supplied by the battery is dissipated as heat during the charging process.

Option D (Statement-1 is false, Statement-2 is true) is correct because electrostatic forces do negative work on positive charges moving from the negative to the positive terminal inside the battery, and the electric field is directed from the positive to the negative electrode.

Option D (Zero) is correct because the electric field is the negative gradient of the potential, and a constant potential implies zero electric field.

Option D is correct. Since the battery is disconnected, the charge remains constant. Increasing the plate separation decreases capacitance and increases potential.