Electric potential and capacitance practice · Mixed difficulty

Option D ($32 \times {10^{ - 4}}Nm$ and $64 \times {10^{ - 4}}J$) is correct because it correctly calculates the maximum torque and work done using the dipole moment and electric field.

Option C (3 cm) is correct because the electric field at the surface of the sphere must be less than or equal to the dielectric strength of the surrounding gas.

Option C (Zero and minimum) is correct because when the dipole moment is aligned with the electric field, the net force is zero and the potential energy is at its minimum.

Option A (Be more on the smaller sphere) is correct because the electric potential is inversely proportional to the radius of the sphere for a given charge.

Option C is correct because the potential of a charged conducting sphere is inversely proportional to its radius, so the smaller sphere will have a higher potential.

Option A ($200 \mu \mathrm{C}$) is correct because the total charge is the sum of the charges on each capacitor, which is the product of the total capacitance and the voltage.

Option C (Assertion is true but reason is false) is correct because the electric field due to a dipole varies inversely as the cube of the distance, not the square.

Option D ($\frac{{{\varepsilon _0}AV_0^2}}{d}$) is correct because it represents the work done to increase the separation, calculated from the change in energy stored in the capacitor.

The correct answer is C (Three in parallel) because capacitors in parallel store the most energy for the same voltage.

Option C (Zero) is correct because the potential energy of a dipole is minimum when it is aligned with the electric field ($\theta = 0$).

Option D ($9 \times {10^{ - 1}}$J) is correct because it is the work done calculated from the change in potential energy when the charges are brought closer.

Option A (${\left( {\frac{{pE}}{I}} \right)^{1/2}}$) is correct because the angular frequency of oscillations of an electric dipole in a uniform electric field is given by this formula.

Option D is correct. The ratio of the equivalent capacitance in series to the equivalent capacitance in parallel is 1:11.

Option C ($\frac{1}{2} \varepsilon_0 \frac{\mathrm{V}^2}{\mathrm{~d}^2}$) is correct because energy density is given by $\frac{1}{2} \varepsilon_0 E^2$ and $E = \frac{V}{d}$.

Option A (Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1) is correct because the Earth is used as a zero potential reference, and the formula for the potential of a charged sphere is correctly stated and relevant to understanding grounding.

Option A is correct because the electric field is the negative gradient of the electric potential, calculated by taking partial derivatives with respect to x, y, and z.

Option D ($\frac{{10}}{9}\mathrm{V}{\mathrm{\mu }^{ - 1}}{\mathrm{m}^{ - 1}}\mathrm{\;and\;in\;the} + \mathrm{ve\;}x\mathrm{\;direction}$) is correct because the electric field is the negative derivative of the potential, and evaluating this at x=4 gives a positive result.

Option C ($\frac{1}{2}\mu F$) is correct because the capacitance is proportional to the radius, and the total volume remains constant when the drop is broken into smaller drops.

Option A ($\mathrm{K}_{\text {air}} < \mathrm{K}_{\text {rubber}} < \mathrm{K}_{\text {copper}}$) is correct because the dielectric constant generally increases from gases (air) to insulators (rubber) to conductors (copper).

Option C ($8kV$) is correct because the maximum voltage is determined by the capacitor that reaches its voltage limit first when connected in series.

Option C ($U$ and $Q$) is correct because when a dielectric is inserted without disconnecting the battery, the voltage remains constant, the capacitance increases, leading to an increase in both charge and stored energy.

Option B is correct. Polar molecules have a permanent electric dipole moment due to the asymmetrical arrangement of atoms and their charges.

Option B is correct because the energy stored in a capacitor is in the form of potential energy due to the electric field between the plates.

Option D is correct because the Earth can accept or donate an almost unlimited amount of charge without significantly changing its potential.

Option A (Lamp will not glow) is correct because a capacitor blocks DC current, and in a series circuit with a capacitor and a lamp connected to a DC source, the lamp will not light up after the capacitor is fully charged.

Option B (Ellipse) is correct because the equipotential surface is defined by $U = constant$, leading to the equation of an ellipse.

Option D (zero) is correct because the work done in moving a charge in an electric field depends only on the initial and final positions, and since the distance between the charges remains the same, the potential energy doesn't change, and thus the work done is zero.

Option D ($\frac{{ - 4{q^2}}}{{\sqrt 3 \pi {\varepsilon _0}b}}$) is correct because it calculates the total potential energy due to the interaction of the central charge with all eight corner charges.

Option A (It measures potential in open circuit) is correct because a potentiometer measures potential without drawing current from the circuit, ensuring an accurate reading.

Option B is correct. The assertion is true because a negative charge moves from low to high potential. The reason is also true because the electric potential energy decreases, and kinetic energy increases, but the reason doesn't directly explain the assertion.