Electrostatics – NEET UG PhysicsPractice Questions & PYQs

Capacitors connected in parallel simply sum their individual values, meaning the parallel branch combines to produce an intermediate capacitance of 6 microfarads. This resulting branch is then wired in series with the remaining 3 microfarads capacitor. Applying the standard network reduction formula for series arrangements, given by $C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2}$ where the two branch values are 3 and 6, the total equivalent capacitance simplifies to 2 microfarads.

The torque experienced by an electric dipole in a uniform electric field is governed by the equation $\tau = p E \sin \theta$. Substituting the given values into this relation gives $4 = p \times 2 \times 10^5 \times \sin(30^{\circ})$, which calculates to a dipole moment of $4 \times 10^{-5}$ C m. Dividing this dipole moment by the dipole length of $0.02$ m yields the magnitude of the charge as $2 \times 10^{-3}$ C, which simplifies to $2$ mC.

The capacitance of a parallel plate capacitor depends directly on the area of the plates and inversely on the distance between them, expressed through the relation $C = \frac{\epsilon_0 A}{d}$. When the plate area is reduced to half its initial value and the separation distance is doubled, the new capacitance becomes $C' = \frac{\epsilon_0 (A/2)}{2d}$. Simplifying this expression yields $C' = \frac{1}{4} \left(\frac{\epsilon_0 A}{d}\right)$, which directly evaluates to $\frac{C}{4}$.

When a charged capacitor is connected to an uncharged capacitor, charge redistributes between them until both share a common potential, following the principle of conservation of total charge. First, the initial charge on the battery-connected capacitor is determined using the relation $q = CV$, yielding $9 \times 10^{-8}$ C for a capacitance of $900$ pF and a voltage of $100$ V. When this capacitor is linked to another identical uncharged capacitor of $900$ pF, the common potential is calculated using the formula $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$, which evaluates to $50$ V. Finally, the total electrostatic energy stored in the two-capacitor system is found using the expression $U = \frac{1}{2}(C_1 + C_2)V^2$, resulting in a stored energy of $2.25 \times 10^{-6}$ J.

The electrostatic potential at the surface of a hollow conducting sphere is determined by the formula $V = \frac{KQ}{R}$, where $Q$ is the charge and $R$ is the radius. When two spheres possess an equal charge, the potential is inversely proportional to their respective radii. Consequently, the smaller sphere, having a significantly smaller radius, develops a higher surface potential than the larger sphere.

The equivalent capacitance for two capacitors connected in series is given by $\frac{C_1 C_2}{C_1 + C_2} = 3$, while their parallel combination gives $C_1 + C_2 = 16$. Substituting the parallel sum into the series equation reveals that the product of the two capacitances is $C_1 C_2 = 48$. Using the algebraic identity $(C_1 - C_2)^2 = (C_1 + C_2)^2 - 4C_1 C_2$, the difference between the capacitances is found to be $8$. Solving this linear system simultaneously yields individual capacitance values of $12 \mu\text{F}$ and $4 \mu\text{F}$.

The work done in moving a charge from infinity to any point is determined by the formula $W = q_0 V$, where $V$ represents the electric potential at the destination. At the centre of the regular hexagon, the alternating distribution of $+q$ and $-q$ charges creates symmetrical electric potentials that are equal in magnitude but opposite in sign. Because every corner charge is equidistant from the centre, these individual potentials completely cancel each other out, resulting in a net electric potential of zero. Consequently, multiplying this zero potential by the transferred charge $q_0$ means no work is required to bring the charge from infinity.

The relationship between potential difference and electric field is expressed by the differential equation $dV = -\vec{E} \cdot d\vec{r}$, which expands in magnitude to $dV = -E \, dr \cos\theta$. On an equipotential surface, the potential is constant everywhere, meaning the change in potential $dV$ is zero. For this dot product to vanish without the field or displacement being zero, the cosine term must be zero, proving that electric field lines are always perpendicular to equipotential surfaces.

An electric dipole consists of two equal and opposite point charges separated by a small fixed distance, creating a localized charge distribution. When calculating the electric field at a distance much larger than this separation, the individual inverse-square fields of the two charges almost completely cancel out due to superposition. This cancellation leaves a net field proportional to the dipole moment divided by the cube of the distance, expressed mathematically as $E \propto \frac{1}{R^{3}}$. Therefore, the field intensity falls off much faster than that of a single point charge.

The potential energy of an electric dipole in an external electric field is determined by the relation $U = -\vec{P} \cdot \vec{E}$, which yields a value of $+PE$ when the dipole is anti-aligned with the field at 180 degrees. Because the surrounding electric field strength decreases toward the right, the dipole experiences a net force that drives it toward regions of lower potential energy while the net torque remains zero. Consequently, the dipole will move toward the right as its potential energy naturally decreases.

The electric potential of a charged spherical drop is expressed by the formula $V = \frac{kq}{r}$, where $q$ is the charge and $r$ is the radius. When twenty-seven identical drops combine, the total volume remains constant, meaning the radius of the new drop scales as the cube root of twenty-seven, yielding a new radius equal to three times the original radius ($R = 3r$). Simultaneously, total charge increases linearly to twenty-seven times the individual charge ($Q = 27q$) due to charge conservation. Substituting these new values into the potential formula gives $V_{\text{big}} = \frac{k(27q)}{3r} = 9\left(\frac{kq}{r}\right)$. Multiplying nine by the initial potential of 220 V results in a final potential of 1980 V.

The energy stored in a capacitor can be determined by multiplying the electrostatic energy density of the electric field by the total volume of the space between the plates. The energy density stored per unit volume is given by $\frac{1}{2} \varepsilon_0 E^2$, and the volume of the region between the plates is equal to the product of plate area and separation distance, $A d$. Multiplying this energy density by the volume yields the total stored energy expression $\frac{1}{2} \varepsilon_0 E^2 A d$.

When a conducting wire directly links the terminals of a component, the resulting equipotential connection short-circuits that element so it no longer stores any charge. By tracking these connected nodes and eliminating the inactive branch, the circuit structure simplifies. The remaining active components are arranged in a parallel configuration across the terminals, meaning their individual values simply add together to yield $C_{AB} = C + C = 2C$.

Connecting two conducting spheres with a wire ensures they reach a common electrostatic potential, meaning $V_1 = V_2$. Expressing potential in terms of charge and radius leads to $\frac{Q_1}{R_1} = \frac{Q_2}{R_2}$, which can be rearranged by introducing surface charge density through the relation $\sigma = \frac{Q}{4\pi R^2}$. This substitution transforms the equality into $\sigma_1 R_1 = \sigma_2 R_2$, producing the inverse ratio of surface charge densities $\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}$.

Polar molecules are characterized by an asymmetric internal charge distribution where the center of positive charge does not coincide with the center of negative charge. This permanent separation of charge centers within the molecular structure gives rise to a permanent electric dipole moment even in the absence of any external electric field. Consequently, these molecules behave like tiny permanent dipoles that naturally possess an intrinsic dipole moment.

The capacitance of a parallel plate capacitor increases upon introducing a dielectric medium according to the relation $C = KC_0$, where $K$ is the dielectric constant of the medium. The dielectric constant is also defined as the ratio of the permittivity of the medium to the permittivity of free space, expressed as $K = \frac{\epsilon}{\epsilon_0}$. Finding the dielectric constant by dividing the new capacitance by the initial air capacitance and multiplying that result by $\epsilon_0$ yields the permittivity of the medium. Computing this product gives a permittivity value of $0.44 \times 10^{-10}$ C$^2$N$^{-1}$m$^{-2}$.

The electric potential at any point around a short electric dipole depends on its dipole moment, distance, and angular orientation relative to the axis, expressed by the formula $V = \frac{kp \cos\theta}{r^2}$. Substituting the electrostatic constant, the dipole moment value of $16 \times 10^{-9}$ C m, the cosine of 60 degrees, and the square of the distance 0.6 m into the expression allows direct computation. Evaluating this product and quotient combination simplifies the value to 200 V.

The electric field outside a charged spherical conductor is calculated using the inverse-square law formula $E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$, treating the entire charge as though it were concentrated at the centre. Substituting the electrostatic constant, the total charge of $3.2 \times 10^{-7}$ C, and the distance of 0.15 m from the centre into the equation yields the magnitude. Evaluating this expression provides the electric field strength of $1.28 \times 10^5$ N/C.

The electric field in any region is fundamentally linked to the spatial gradient of the electric potential via the differential relation $E = -\frac{dV}{dr}$. Since the electric potential remains constant at 5 V throughout the entire volume, its derivative with respect to position is zero. Consequently, the absence of any potential gradient dictates that the electric field in this region is zero.

Electrostatic force between two stationary point charges is governed by Coulomb's law, which relates the force directly to the product of their charges and inversely to the square of the separation distance. Initially, the charges are $+Q$ and $-Q$, producing a baseline attractive force $F$. When 25% of the charge from the first body is shifted to the second, the respective charges become $\frac{3Q}{4}$ and $-\frac{3Q}{4}$. Substituting these updated values into Coulomb's equation yields a new force magnitude of $\frac{9}{16} \frac{k Q^2}{r^2}$, which simplifies directly to $\frac{9F}{16}$.