Electrostatics – NEET UG PhysicsPractice Questions & PYQs

The electric field is related to the electric potential by the negative gradient, represented mathematically as $\vec{E} = -\nabla V$. To find each directional component, we compute the partial derivatives of the potential function $V = -x^2y - xz^3 + 4$ with respect to each spatial coordinate.

Differentiating partially with respect to $x$ gives $-2xy - z^3$, which reverses in sign to yield $2xy + z^3$ for the first component. Taking the partial derivative with respect to $y$ produces $-x^2$, resulting in $x^2$ after applying the negative sign. Finally, the partial derivative with respect to $z$ evaluates to $-3xz^2$, leading to $3xz^2$ for the final component, which altogether assembles into the complete field expression $\vec{E} = \hat{i}(2xy+z^3)+\hat{j}x^2+\hat{k}(3xz^2)$.

Equivalent capacitance in a series combination is found by taking the reciprocal sum of the individual capacitances, which for three identical capacitors reduces the total capacitance to one-third of the individual value. Because identical capacitors in series share the applied voltage equally, each individual unit reaches its individual breakdown limit only when the total external voltage reaches three times that limit. Consequently, the combination can safely sustain a threefold increase in total voltage before any single component fails.

The net electric field at the centre of a uniformly charged closed ring is always zero, establishing the vector relation:

$\vec{E}_{AKB} + \vec{E}_{ACDB} = 0$

By evaluating the symmetry of the remaining portion of the ring, the field contributions from the lateral side segments cancel each other out entirely at the centre. Consequently, the opposite segment generates a field with a magnitude equal to $E$ that points in the reverse direction along $OK$ to satisfy this zero-field condition.

The energy stored in a charged parallel plate capacitor can be determined by integrating the electrostatic energy density throughout the volume occupied by the electric field. The energy density stored in a uniform electric field is given by $\frac{1}{2}\varepsilon_0 E^2$. Multiplying this energy density by the total volume between the plates, which is the product of area and separation distance, yields the total required charging energy.

The electric potential of a point charge is expressed as $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$, and the electric field magnitude is given by $E = \frac{V}{r}$. Equating the given potential value to the standard formula allows the reciprocal distance term to be expressed as $4\pi\varepsilon_0 \times 10^{11}$. Multiplying this result by the given potential value directly provides the required electric field magnitude.

When capacitors are connected in a parallel configuration across a voltage source, the total work done by the battery in charging them fully is determined by the relation $W = \frac{1}{2}QV$, where $Q$ is the total charge delivered to the circuit. Since the condensers are arranged in parallel, the potential difference across both individual components matches the battery voltage $V$. Therefore, the charge accumulated on the first capacitor is $Q_1 = \mathrm{C}V$ and the charge on the second capacitor is $Q_2 = \frac{\mathrm{C}}{2}V$. Adding these quantities together gives the total system charge of $Q = \mathrm{C}V + \frac{\mathrm{C}}{2}V = \frac{3}{2}\mathrm{C}V$. Substituting this total charge back into the work equation produces $W = \frac{1}{2} \left( \frac{3}{2}\mathrm{C}V \right) V$, which simplifies neatly to $\frac{3}{4}\mathrm{C}V^2$.

Gauss's law establishes that the total electric flux through any closed boundary depends solely on the net enclosed charge divided by the permittivity of free space, given by $\frac{q}{\varepsilon_{0}}$. For a closed hollow cylinder, this total flux is distributed across three distinct boundaries: the curved lateral surface and the two identical flat circular end faces. Because of the symmetrical geometry, the electric flux passing through the flat end surface $A$ matches the flux escaping through the opposite flat end. Therefore, the sum of the flux through the curved surface $B$ and twice the flux through surface $A$ equals the total enclosed flux of $\frac{q}{\varepsilon_{0}}$. Rearranging this relation by substituting the known curved surface flux $\phi$ and isolating the flux for surface $A$ results in $\frac{1}{2}\left(\frac{q}{\varepsilon_{0}}-\phi\right)$.

The central charge can be conceptualized as two separate negative charges, allowing the system to be split into two individual electric dipoles oriented along the coordinate axes. One dipole consists of the charges located along the y-axis with a dipole moment of $q a$ directed along the positive y direction. The second dipole consists of the charges located along the x-axis with an equal magnitude of $q a$ directed along the positive x direction. Combining these two perpendicular dipole vectors using vector addition yields a net magnitude of $\sqrt{q^2 a^2 + q^2 a^2} = \sqrt{2} q a$. The resultant vector points along the angle bisector between the axes, which corresponds to the line joining the origin to the point having equal positive coordinates.

The work required to move a charge in an electrostatic field is determined by the equation $W = Q(V_D - V_C)$ where $Q$ is the moving charge, and $V_C$ and $V_D$ are the electric potentials at the start and end points. The electric potential at the midpoint $C$ evaluates to zero because the positive and negative source charges are separated by a distance of $2L$ and maintain equal, opposing influences at this exact center. Moving along the semicircle to point $D$, the distance from the charge at $A$ becomes $3L$ while the distance from the charge at $B$ is $L$, establishing a net potential of $-q / (6 \pi \varepsilon_{0} L)$ at this location. Substituting these potential values into the work equation yields a total work done of $-\frac{qQ}{6\pi\varepsilon_0 L}$.

The work required to rotate a dipole in an electric field is determined by integrating the torque $\tau = p E \sin \theta$ from the initial alignment angle to the final angle. Evaluating this integral yields the work formula $W = p E (\cos \theta_1 - \cos \theta_2)$. Substituting the initial angle of zero degrees and the final angle of 90 degrees into this expression simplifies the calculation. Since the cosine of zero degrees is one and the cosine of 90 degrees is zero, the total work done simplifies directly to $p E$.

When the charging battery is disconnected, the total charge stored on the capacitor plates remains completely constant. Increasing the distance between the plates decreases the capacitance because capacitance is inversely proportional to plate separation. Since the potential difference is equal to the ratio of the fixed charge to the reduced capacitance, expanding the gap between the plates directly causes the potential difference to increase.

Electric flux is determined by the integral relation $\Phi_{E} = \int \vec{E} \cdot d\vec{S}$. Since the square surface and the uniform electric field both lie flat in the plane of the paper, the normal vector of the surface forms an angle of 90 degrees with the field. Because the scalar product evaluates to zero when the angle between the vectors is 90 degrees, no field lines cross the surface, yielding zero net flux.

Electrostatic force is a conservative force, meaning the work done in moving a charge depends solely on the change in potential energy between the initial and final positions. The electric potential due to a point charge is expressed as $V = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r}$, showing it depends entirely on the radial distance $r$ from the origin. Because point $A$ located at $(0, a)$ and point $B$ located at $(a, 0)$ are both at a distance $a$ from the source charge $+q$, their electric potentials are identical. Since moving the charge $-Q$ between these two points involves no change \in potential energy, the work done along the path $AB$ is zero.

The total electrostatic potential energy of a system of charges is determined by summing the pairwise potential energy interactions across all charge pairs. When the charge $q_3$ resides at point $C$, its initial potential energy $U_1$ consists of the interaction with $q_1$ at a distance of $0.40$ meters and with $q_2$ at a distance of $\sqrt{0.40^2 + 0.30^2} = 0.50$ meters. Upon moving $q_3$ along the arc to point $D$, its separation from $q_1$ stays at $0.40$ meters, whereas its separation from $q_2$ decreases to $0.40 - 0.30 = 0.10$ meters, defining the final potential energy $U_2$. Finding the difference $\Delta U = U_2 - U_1$ causes the terms associated with $q_1$ to cancel out identically. Evaluating the remaining terms yields $\Delta U = \frac{q_3}{4 \pi \varepsilon_0} \left( \frac{q_2}{0.10} - \frac{q_2}{0.50} \right)$, which simplifies to $\frac{q_3}{4 \pi \varepsilon_0} (8 q_2)$, thereby establishing that $k$ is $8 q_2$.

Capacitors connected in a series branch share the exact same charge, and their collective behavior is determined by the reciprocal capacitance formula. For the branch containing $C_1$, $C_2$, and $C_3$, the equivalent capacitance $C'$ is calculated as:

$\frac{1}{C'} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{3C} = \frac{11}{6C}$

This yields an equivalent capacitance of $C' = \frac{6C}{11}$. Because these three elements are wired in series, the charge residing on $C_2$ matches the total charge of this branch, which evaluates to $Q' = \frac{6}{11}CV$. On the other hand, the fourth capacitor $C_4$ operates independently in a parallel branch across the same voltage, accumulating a charge of $Q = 4CV$. Dividing the charge of $C_2$ by the charge of $C_4$ gives the final ratio of $\frac{3}{22}$.

When an electric dipole is placed in a uniform electric field, each equal and opposite charge experiences a force of identical magnitude pointing in opposite directions, causing the net translational force on the dipole to vanish completely. The potential energy of the dipole in this field is governed by the orientation of its dipole moment relative to the field lines, achieving its lowest value when both vectors point in the identical direction. Consequently, since the dipole moment is aligned directly along the field, the system configuration reaches minimum potential energy.

The kinetic energy acquired by the accelerated charge is supplied entirely by the electrical work done, establishing the relation $\frac{1}{2} m v^2 = q V$. Converting the mass and charge into standard units allows us to set up the calculation for the required electric potential. Rearranging the energy equation to solve for potential yields $V = \frac{m v^2}{2 q}$. Substituting the given values into this expression results in a potential difference of 50 kV.

Gauss's law establishes that the total electric flux emanating from a closed surface enclosing a charge is proportional to the enclosed charge divided by the permittivity of free space. Since a cube consists of six identical faces symmetrically surrounding the center, this total flux is distributed equally across all faces. Dividing the total flux by six yields the flux through a single face, which can be algebraically rearranged to match the required format by multiplying and dividing by $4\pi$. Thus, the expression becomes $\phi_{\text{face}} = \frac{4 \pi q}{6 (4 \pi \varepsilon_{0} )}$.

The equivalent capacitance of two identical devices connected in series is halved, while a parallel arrangement sums their individual capacities. Pairing two of the 4 µF capacitors in series yields an intermediate value of 2 µF. Connecting the remaining 4 µF capacitor in parallel with this pair gives a final equivalent value of $C_{\text{eq}} = \frac{4 \times 4}{4+4} + 4$, which simplifies to $2 + 4 = 6$ µF.

When an initially charged capacitor is connected to an uncharged capacitor, charge redistributes until a common potential is reached across both devices. The total charge residing in the system is determined entirely by the initial capacitance and voltage, expressed as $q = C_{1} V$. Because the capacitors are now in parallel, their individual capacitances add up to a combined equivalent capacitance of $C_{1} + C_{2}$. Dividing the conserved total charge by this new total capacitance yields the final potential difference as $V' = \frac{C_{1} V}{C_{1}+C_{2}}$.