Electrostatics – NEET UG PhysicsPractice Questions & PYQs

The electrostatic energy stored in any capacitor is given by $U = \frac{1}{2} C V^2$, where $V$ represents the potential difference across that specific component. In the given circuit, capacitors $P$ and $Q$ are connected in series across the power source, causing the battery voltage to divide evenly so that the potential across capacitor $P$ is $V/2$. At the same time, capacitor $S$ is linked directly across the supply terminals and experiences the full potential of $V$.

Applying the energy formula to each element results in $U_P = \frac{1}{2} C (V/2)^2 = \frac{C V^2}{8}$ for capacitor $P$, $U_Q = \frac{C V^2}{8}$ for capacitor $Q$, and $U_S = \frac{1}{2} C V^2$ for capacitor $S$. Combining these individual amounts determines the total system energy $U_T = \frac{3}{4} C V^2$. Dividing the energy of capacitor $P$ by this total system energy yields a ratio of $1/6$.

When two separate metallic spheres are connected by a conducting wire, charge flows between them until both bodies reach a common electrostatic potential, calculated using the relation $V = \frac{Q_1 + Q_2}{4\pi\varepsilon_0 (R_1 + R_2)}$. This shared potential is determined by dividing the total initial charge by the sum of their individual capacitances. Substituting the given radii and initial charges into this framework yields the unified potential, which is then multiplied by the capacitance of the smaller sphere to determine its final retained charge of $2 \times 10^{-2}$ C.

The electrostatic potential energy of an electric dipole immersed in a uniform electric field is determined by the orientation of the dipole relative to the field lines, expressed as $U = -pE\cos\phi$. Stable equilibrium corresponds to an initial alignment where the dipole moment points directly along the field, making the angle $\phi$ equal to zero. Rotating the dipole through an angle $\theta$ changes this orientation angle, resulting in a final potential energy of $-pE\cos\theta$.

The electrostatic potential energy stored in a capacitor is accumulated as work is done to transfer charge between its plates, expressed by the foundational relation $U = \frac{1}{2} C V^2$. As charge builds up on the capacitor, the potential difference increases proportionally, meaning each subsequent increment requires more work to move against the existing field. Integrating this incremental work yields the final energy expression in terms of capacitance and voltage.

The total mechanical energy of the system remains conserved when electrostatic forces act alone, meaning any reduction in potential energy transforms entirely into kinetic energy. The initial electrostatic potential energy when the charge is located at a distance of $3R$ is determined by the Coulomb interaction, and the final potential energy is evaluated at the surface radius $R$. Subtracting the final potential energy from the initial potential energy yields the net drop in potential energy as $\Delta U = \frac{2 kQq}{3R}$. Equating this energy loss to the kinetic energy expression $\frac{1}{2}mv^2$ and substituting the Coulomb constant $k = \frac{1}{4\pi\varepsilon_0}$ allows solving directly for the final velocity.

The electrostatic energy lost when a charged capacitor connects to an identical uncharged capacitor is derived from the initial stored energy and the conservation of charge. Using the standard energy loss formula for parallel connection of two capacitors of equal capacitance and initial voltage, the dissipated energy is given by $\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} V^2$. Substituting the capacitance value of $200$ pF and the initial voltage of $100$ V simplifies the expression. Evaluating this numerical expression yields the final energy loss of $0.5 \times 10^{-6}$ J.

A charge is in stable equilibrium when the net electrostatic force on it is zero and any small displacement generates a restoring force pushing it back toward its original position. Evaluating the forces on the central negative charge reveals that the attractive pulls from the two surrounding positive charges exactly cancel each other out. If this negative charge is slightly displaced sideways, the closer positive charge exerts a stronger pull, creating a restoring force that drives it back to the center while leaving the outer charges unstable.

Finding the equivalent capacitance of a complex network involves systematically reducing series and parallel sections, where parallel branches add directly according to $C_{\text{eq}} = C_1 + C_2$ and series branches combine via reciprocal summation. Tracing the circuit from the terminals inward allows us to identify symmetrical groupings and simplify local loops into single equivalent components. Continuing this stepwise reduction across the remaining junctions collapses the entire web of capacitors into a final equivalent capacitance of 6 µF.

The electric field produced by an electric dipole at a distant point on its perpendicular bisector is governed by the formula $E = \frac{p}{4\pi\varepsilon_0 r^3}$ under the condition that the distance is much larger than the dipole length. Examining this relationship reveals that the field magnitude is directly proportional to the dipole moment $p$ and inversely proportional to the cube of the distance $r$.

The work done in moving a charge slowly between two points in an electrostatic field is equal to the product of the charge and the potential difference between those points, expressed as $W = q(V_B - V_A)$. Since both the initial position at $2R$ and the final position at $3R$ lie outside the uniformly charged dielectric sphere, the electric potential at a distance $r$ from the center behaves like a point charge and is given by $V = \frac{Q}{4\pi\varepsilon_0 r}$.

For a unit positive point charge moving between these locations, the magnitude of the work done is evaluated by substituting the respective radial distances into the potential expression, resulting in $W = \frac{Q}{4\pi\varepsilon_0} \left( \frac{1}{3R} - \frac{1}{2R} \right)$, which simplifies in magnitude to $\frac{Q}{24\pi\varepsilon_0 R}$.

Relating the total charge of the sphere to its volume and uniform charge density via $Q = \rho \cdot \frac{4}{3}\pi R^3$ and substituting this back into the work equation yields $\frac{\rho R^2}{18\varepsilon_0}$. Matching this resulting expression to the given form $\frac{\rho R^2}{n \varepsilon_0}$ directly establishes that the value of $n$ is $18$.

Electrostatic shielding dictates that the electric field inside any closed conducting cavity or hollow conductor is strictly zero. Because any excess charge given to a conductor resides entirely on its outer surface, no electric field lines can penetrate the interior space. Even though the hollow sphere possesses a positive charge and a specific radius, the electric field at its exact centre remains completely zero.

For a multi-particle charge system to remain in complete equilibrium, the net electrostatic force acting on every individual charge must vanish. Let the two charges $Q$ be separated by a distance $2r$, placing the central charge $q$ at a distance $r$ from each. Evaluating the forces on one of the outer charges $Q$, the repulsive force exerted by the other charge $Q$ is balanced by the attractive force from the central charge $q$. Equating the sum of these forces to zero gives $\frac{Q^2}{(2r)^2} + \frac{qQ}{r^2} = 0$, which simplifies to $q = -\frac{Q}{4}$.

A conductor in electrostatic equilibrium maintains a zero internal electric field because mobile charges rearrange until internal forces vanish. Any net excess charge given to the conductor resides entirely on its outer surface, leaving the interior completely neutral. Furthermore, the electric field lines at the conductor surface must always intersect perpendicularly to ensure no tangential component of the electric field exists.

Gauss's law states that the total electric flux passing through any closed surface is determined entirely by the net charge enclosed within that surface, expressed as $\Phi = \frac{q}{\varepsilon_0}$. Because the point charge is situated at the exact center, the enclosed charge dictates the entire outgoing flux independently of the cube's side length. Thus, the physical dimensions of the cube have no bearing on the final flux value.

Like charges repel each other naturally, meaning external work must be performed to push them closer together against their mutual electrostatic repulsion. Since the energy supplied during this mechanical compression is stored within the system configuration, the electrostatic potential energy accumulates. Consequently, reducing the separation between two electrons causes their mutual potential energy to steadily increase.

Finding the equivalent capacitance of a mixed circuit relies on combining the series formula $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_4}$ with the parallel summation formula $C_{\text{eq}} = C_s + C_5$. The four identical $10\ \mu\text{F}$ capacitors are wired in series, which simplifies their combined value to $2.5\ \mu\text{F}$. This series group is then connected in parallel with the fifth capacitor of $2.5\ \mu\text{F}$, bringing the overall equivalent capacitance of the network to $5\ \mu\text{F}$ when hooked up to a $50\text{ V}$ battery.

Because both parallel branches experience the full source potential, the charge coursing through the series branch is determined using $q = C_s \times V$, resulting in $125\ \mu\text{C}$ that passes through every single capacitor in that line. At the same time, the charge on the fifth capacitor is found by multiplying its capacitance of $2.5\ \mu\text{F}$ by the $50\text{ V}$ voltage, which also evaluates to $125\ \mu\text{C}$. Consequently, every single component in the entire arrangement holds an identical charge alongside the $5\ \mu\text{F}$ total capacitance.

Coulomb's law establishes that the electrostatic force between charges is inversely proportional to the dielectric constant of the intervening medium. When air or vacuum is replaced by a dielectric medium, the effective force is scaled down by a factor of $K$. Therefore, the new force becomes $F_{m} = \frac{F_{0}}{K}$, indicating that it decreases $K$ times.

The work done in moving a charge along an equipotential path is zero because the electric potential remains constant at every point along the trajectory. Since work depends directly on the potential difference between the initial and final positions, completing a full circuit back to the starting point means the net potential difference is zero. Consequently, multiplying this zero potential difference by the moving charge yields no work done.

The work-energy theorem states that the work done by an electric field on a charged particle is entirely converted into its kinetic energy. The electric field exerts a constant force $F = qE$, which generates an acceleration of $a = \frac{qE}{m}$. Substituting this acceleration into the kinematic equation $v^{2} = 2ay$ and evaluating the kinetic energy expression $\text{K.E.} = \frac{1}{2}mv^{2}$ successfully simplifies to yield $qEy$.

Inside a hollow metallic conductor in electrostatic equilibrium, the electric field is strictly zero, which maintains a uniform electric potential throughout the interior volume. Because the potential gradient vanishes where the electric field is zero, the electric potential at any internal point equals the potential found at the outer surface. Therefore, the potential at the center matches the surface value of 80 V.