NEET Electrostatic Potential and Capacitance Previous Year Questions with Solutions

Capacitors P and Q are connected in series across the battery voltage V. Since they have identical capacitance C, the battery voltage divides equally across them:

• • Capacitor Q also stores exactly U_Q = (1/8) * C * V^2.

Capacitor S is connected in parallel directly across the battery terminals, experiencing the full potential difference V:

• • Total energy stored in the entire system is the sum of energies in all three capacitors:
• • U_T = U_P + U_Q + U_S = \frac{1}{8}CV^2 + \frac{1}{8}CV^2 + \frac{4}{8}CV^2 = \frac{6}{8}CV^2 = \frac{3}{4}CV^2.

Divide the energy stored in P by the total system energy:

• • The factor C * V^2 cancels cleanly.
• • Leaves the exact fraction 1/6, matching Option (4).

By Gauss's Law, outside a uniformly charged dielectric sphere of radius R and charge density rho, the electric field and potential are identical to a point charge Q at the origin:

• • Because both point A (r = 2R) and point B (r = 3R) lie outside the sphere, this potential formula applies directly to both endpoints.

Substitute r_A = 2R and r_B = 3R into the exterior potential formula:

• • Common denominator between 6 and 9 is 18.

Because the electrostatic field is conservative, work done on unit positive charge is given by the potential difference:

• • Comparing with the given expression |W| = \frac{\rho R^2}{n \epsilon_0}:
• • n = 18.
• • Matches Option (4) exactly.

Point A is located in the free space inside the cavity containing point charge Q:

• • Electric field is zero inside the conducting METAL body, but NOT inside an empty cavity containing an enclosed charge.
• • Point charge Q produces a strong radial field in its vicinity, so E_A is non-zero.

Apply Gauss's law to a Gaussian surface enclosing the cavity inside the conductor:

• • By charge conservation on the isolated conductor: q_{outer} = +Q.
• • Because the conductor's exterior surface is a sphere, electrostatic equilibrium forces +Q to distribute with uniform surface charge density sigma = Q / (4pi R_{ext}^2).

By Gauss's theorem, outside a uniformly charged spherical surface, the field is purely radial from the center O:

• • The off-center position of the cavity has zero effect on the external field symmetry.
• • Therefore, E_A != 0 and E_B = E_C.
• • Matches Option (2) exactly.

Because the insulating sphere is fixed and electrostatic forces are conservative:

• • Initial kinetic energy K_i = 0 (released from rest).
• • Initial distance r_i = 3R; final distance r_f = R.

Outside a uniformly charged sphere, electrostatic potential energy between +Q and -q is:

• • At r = 3R: U(3R) = -\frac{Qq}{12\pi \epsilon_0 R}.
• • At r = R: U(R) = -\frac{Qq}{4\pi \epsilon_0 R}.

Isolate (1/2)*m*v^2 and simplify:

• • Multiply by 2/m: v^2 = \frac{2 Qq}{6\pi \epsilon_0 m R} = \frac{Qq}{3\pi \epsilon_0 m R}.
• • Taking the square root gives: v = \sqrt{\frac{Qq}{3\pi \epsilon_0 m R}}.
• • Matches Option (3) exactly.

For a parallel plate capacitor of area A and plate separation d containing dielectric slabs in series, the capacitance is:

• • t₁ = (3/8) d, t₂ = (1/2) d = (4/8) d.
• • Air gap thickness: t_air = d - (3/8 d + 4/8 d) = (1/8) d.

The problem states that C = 2 C₀, where C₀ = ε₀ A / d:

• • Subtract 1/8 from both sides: 3/(8 K₁) + 4/(8 K₂) = 1/2 - 1/8 = 3/8.
• • Multiplying throughout by 8 gives the simplified linear form: 3/K₁ + 4/K₂ = 3.

Given K₁ = 1.25 K₂ = (5/4) K₂, which implies K₂ = (4/5) K₁:

• • Matches Option (1) with absolute precision.

The given circuit consists of five capacitors each having capacitance C = 2 μF arranged in a Wheatstone bridge configuration. The condition for balance in a capacitance bridge is:

Because the ratio of adjacent arms is equal, the electric potential at the upper junction and lower junction is identical. No charge flows into or through the central 2 μF capacitor, so it can be completely removed from the circuit without affecting the equivalent capacitance.

With the central capacitor removed, the circuit simplifies to two parallel branches, each containing two 2 μF capacitors in series:

The top and bottom branches are connected in parallel across terminals A and B. Their capacitances add directly:

Consider a Gaussian spherical surface of radius r < R concentric with the shell. Because all charge q resides on the outer surface of the thin shell, the charge enclosed within the Gaussian sphere is zero. By Gauss’s law:

The electric field is the negative gradient of the electrostatic potential:

Because potential is constant everywhere from the center up to the surface, the potential at the center C is identically equal to the potential at point P on the surface:

The potential difference between points C and P is therefore identically zero:

Because the capacitor remains connected across the terminals of a battery throughout the operation, the potential difference V between its plates is held strictly constant by the battery EMF:

Capacitance of a parallel plate capacitor with plate area A and separation d is given by C = ε₀A / d. When the plates are moved closer together, separation d decreases, which directly increases capacitance:

The charge Q stored on the plates is Q = C · V. With V held constant and C increasing, the battery supplies additional charge to the plates:

The electrostatic potential energy U stored in the electric field of the capacitor is U = ½ C V². Since C increases and V is constant, stored energy increases:

The ratio of charge to potential is the capacitance itself (Q/V = C). Since capacitance increases, the ratio changes and does not remain the same:

The product of charge and voltage is Q · V = (CV) · V = C V². Because C increases and V is constant, the product increases:

Inspect the right branch of the circuit: two identical capacitors of capacitance 3 μF are connected in parallel across the same potential nodes. The equivalent capacitance of a parallel combination is the direct sum:

The parallel block C_p = 6 μF is connected in series with the first capacitor C_1 = 3 μF between terminals A and B. The equivalent series capacitance is given by the reciprocal formula:

Inverting yields the net equivalent capacitance:

The electric dipole consists of charges -q and +q placed symmetrically along the x-axis around the origin 0: charge -q is at x = -3 cm, and charge +q is at x = +3 cm. Point P lies on the positive x-axis at x = +5 cm.

Calculate the distances r₊ and r₋ from charges +q and -q to point P in meters:

By the principle of superposition, the net electrostatic potential V at point P is the algebraic scalar sum of the potentials due to both individual point charges:

Factor out 10² from the denominator into the numerator:

The question asks for the potential expressed in units of 10² V:

When two conducting spheres of radii $R_1$ and $R_2$ are joined by a thin conducting wire, free electrons flow between them until they reach an identical electrostatic potential $V$:

Surface charge density $\sigma$ is defined as the total charge per unit spherical surface area $A = 4\pi R^2$:

Substitute the charge ratio $q_1 / q_2 = R_1 / R_2$ into the density ratio:

In a molecule, the effective centers of positive charges (atomic nuclei) and negative charges (electron clouds) determine molecular polarity:

NCERT defines both categories explicitly: - Non-polar molecule: The centers of positive and negative charges coincide. Thus, the molecule has no permanent (or intrinsic) dipole moment (e.g., $\text{O}_2, \text{H}_2, \text{CO}_2$). - Polar molecule: The centers of positive and negative charges are separated by a finite displacement even when no external field is applied. Therefore, polar molecules possess an intrinsic, permanent electric dipole moment (e.g., $\text{HCl}, \text{H}_2\text{O}$).

When placed in an external electric field, the permanent dipoles experience a torque $\vec{\tau} = \vec{p} \times \vec{E}$ that tends to align them parallel to the field direction, creating orientational polarization.

The electrostatic potential energy $U$ stored in a charged capacitor of capacitance $C$ with potential difference $V_0$ across its plates is:

For an ideal parallel plate capacitor in vacuum: - Capacitance: $C = \frac{\varepsilon_0 A}{d}$ - Potential difference: $V_0 = E \cdot d$

Canceling one factor of plate separation $d$:

Assign electrostatic potentials to all wire junctions in the circuit diagram: - Let the input terminal on the left be at potential $V_A$. - Let the output terminal on the right be at potential $V_B$. - The ideal continuous conducting wire on the right connects the top node and bottom node directly to terminal B with zero resistance. Therefore, both nodes are at potential $V_B$.

The vertical capacitor of capacitance $C$ has its upper terminal connected to the upper wire (potential $V_B$) and its lower terminal connected to the lower wire (potential $V_B$):

Since the vertical capacitor carries zero charge and contributes zero capacitance, it can be removed from the circuit. The remaining circuit consists solely of the top capacitor and the bottom capacitor connected in parallel between node A and node B:

Let $r$ be the radius of each small drop and $R$ be the radius of the combined bigger drop. Since the liquid is incompressible, the total volume is conserved:

Let $q$ be the charge on each small drop. Total charge is conserved upon coalescence:

The potential of a single small conducting droplet of radius $r$ and charge $q$ is $v = \frac{kq}{r} = 220\text{ V}$. The potential of the big drop is:

Substitute $v = 220\text{ V}$:

For a spherical conductor of radius $$R$$ carrying total charge $$Q$$ uniformly distributed on its surface: - Inside the conductor ($$r < R$$): $$E = 0$$. - Outside the conductor ($$r \ge R$$): By spherical symmetry and Gauss's law, the electric field is identical to that of a point charge $$Q$$ located at the center: $$E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2}$$

The point is at a distance $$r = 15\text{ cm}$$ from the centre. Because $$r = 15\text{ cm} > R = 10\text{ cm}$$, the point lies outside the sphere. $$r = 15\text{ cm} = 0.15\text{ m} = \frac{3}{20}\text{ m}$$

$$E = \frac{(9 \times 10^9\text{ N m}^2\text{ C}^{-2}) \times (3.2 \times 10^{-7}\text{ C})}{(0.15\text{ m})^2}$$ $$E = \frac{28.8 \times 10^2}{0.0225} = \frac{2880}{0.0225}$$ Multiply numerator and denominator by 400/9: $$E = 2880 \times \frac{400}{9} = 320 \times 400 = 128000\text{ N C}^{-1} = 1.28 \times 10^5\text{ N C}^{-1}$$

Comparing with options, $$1.28 \times 10^5\text{ N/C}$$ is uniquely Option (2).

The electric field vector in electrostatics is equal to the negative spatial gradient of the electric potential scalar function: $$\vec{E} = -\vec{\nabla} V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)$$ Along any spatial direction $$r$$: $$E = -\frac{dV}{dr}$$

The problem explicitly states that the potential is found to be $$5\text{ V}$$ throughout the entire volume of $$0.2\text{ m}^3$$. This means $$V(x, y, z) = \text{constant} = 5\text{ V}$$. The derivative of any constant with respect to space coordinates is identically zero: $$\frac{\partial V}{\partial x} = 0, \quad \frac{\partial V}{\partial y} = 0, \quad \frac{\partial V}{\partial z} = 0$$

$$E = \sqrt{0^2 + 0^2 + 0^2} = 0\text{ N C}^{-1}$$ Thus, the magnitude of the electric field in this entire region is zero.

This is identical to the interior of a charged conductor in electrostatic equilibrium: the potential is uniform and constant throughout the interior, and the electric field inside is strictly zero.