Electrostatics – NEET UG PhysicsPractice Questions & PYQs

A point charge placed inside the cavity of a solid isolated conductor experiences no electrostatic shielding inside the hollow space, ensuring that the electric field at point $A$ is not zero so $E_A \neq 0$. Beyond the outer surface of the conductor, the electric field behaves symmetrically as if the net charge resides entirely at the geometric center of the sphere. Because points $B$ and $C$ are situated at identical distances from this center, the field magnitudes at these two locations are equal, confirming that $E_B = E_C$.

Electric field lines fundamentally originate on positive charges and terminate on negative charges. When examining the directional arrows in the given electric field diagram, the lines point inward toward both $q_1$ and $q_2$. Because electric lines always terminate on negative sinks, this inward convergence proves that both are negative charges.

The electrostatic potential energy of a dipole in a uniform electric field is expressed as $U = -pE\cos\theta$, where $\theta$ is the angle between the dipole moment and the electric field. The change in potential energy when the dipole rotates from an initial angle of zero degrees to a final angle of sixty degrees is $\Delta U = -pE\cos\theta_f - (-pE\cos\theta_i)$. Substituting the given dipole moment, electric field magnitude, and cosine values evaluates the net energy difference as exactly one joule.

Two identical spheres initially carrying equal charges experience an initial electrostatic force proportional to the square of their charge. When a third identical uncharged sphere touches the first sphere, the total charge divides equally, leaving both with half of their original charge. Touching this third sphere to the second sphere subsequently redistributes the remaining charge evenly between them, resulting in a new charge product that alters the mutual electrostatic force. Calculating the ratio of the final force to the initial force using Coulomb's law yields a reduced force equal to three-eighths of the original value.

The equivalent capacitance of a capacitor containing multiple dielectric slabs is determined by treating the system as a combination of smaller capacitors connected in series, leading to the general formula $C_{\text{eq}} = \frac{\varepsilon_0 A}{\frac{t_1}{K_1} + \frac{t_2}{K_2} + \frac{t_3}{K_3}}.$ The total plate separation is divided among the two slabs and the remaining air gap, where the third thickness has a dielectric constant of unity. Substituting the given thickness values, the relationship $K_2 = \frac{K_1}{1.25}$, and setting the equivalent capacitance equal to twice the initial capacitance value allows the equation to be solved directly for $K_1$, resulting \in $2.66$.

This circuit functions as a balanced Wheatstone bridge, meaning the electric potential across the central branch is balanced and no charge flows through that middle component. Removing this inactive branch leaves two parallel arms connected between terminal A and terminal B, with each arm simplifying to an equivalent value of 1 microfarad. Combining these two identical parallel branches together gives the final equivalent capacitance of $C_{AB}$ equal to 2 microfarads.

The electric potential at a radial distance from a point charge is determined by the formula $V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$. Substituting the given electrostatic constant, the magnitude of the point charge, and the distance converted into meters, the terms simplify directly. The numerical factors cancel out appropriately, leaving the potential value to evaluate to $4 \times 10^4$ volts.

When a circuit reaches steady state, a fully charged capacitor prevents any charge from flowing through its branch, effectively turning it into an open circuit. Removing this inactive branch leaves a straightforward loop where the total resistance is the sum of the active resistors, giving 5 ohms. Applying Ohm's law through the formula $i = \frac{V}{R}$ divides the source voltage of 10 volts by this total resistance, yielding a steady-state current of 2 A.

The continuity of total current through any circuit containing a charging capacitor is maintained by the introduction of displacement current within the dielectric gap between the plates. As conduction current travels through the connecting wires to charge the plates, a time-varying electric field develops across the gap. According to Maxwell's extension of Ampere's law, this changing electric flux generates a displacement current whose magnitude precisely equals the conduction current in the wire and flows in the same continuous direction.

When a parallel plate capacitor remains connected to a constant voltage source while its plates are brought closer together, its capacitance increases inversely with the separation distance according to $C = \frac{\epsilon_0 A}{d}$. Because capacitance becomes larger while the potential difference stays fixed, the stored charge increases proportionally through $Q = CV$, which also increases both the stored electrical energy and the product of charge and voltage. Consequently, the charge increases, capacitance increases, and the charge-voltage product increases, while the ratio of charge to potential does not remain constant.

The electric potential due to a short electric dipole along its axis is given by the formula $V = \pm \frac{P}{4\pi\epsilon_0 r^2}$. Substituting the dipole moment of $4 \times 10^{-6}\text{ C m}$ and distance of $2\text{ m}$ into this expression results in a potential magnitude of $9 \times 10^3\text{ V}$, confirming that Assertion A is true. However, Reason R incorrectly includes a factor of two in the numerator of its stated formula, making the reason statement false.

The electrostatic energy stored in a charged capacitor is given by the formula $U = \frac{1}{2} C V^2$. Substituting the capacitance value in farads and the potential difference in volts, the expression becomes a straightforward product of numerical values and powers of ten. Squaring the potential and multiplying by half the capacitance yields a total stored energy of $15$ nanojoules.

Surface charge density is defined as the charge per unit surface area, expressed as the formula $\sigma = \frac{Q}{S}$, where all excess charge resides entirely on the outer surface of a conductor. For a metal cube, the total surface area comprises six identical square faces, given by the expression $S = 6a^2$. Substituting the side length converted into meters, the calculated surface area becomes $150 \times 10^{-4}$ square meters. Dividing the total charge of $6$ microcoulombs by this area yields a surface charge density of $0.4 \times 10^{-3}$ coulombs per square meter.

The electric potential at any point inside a uniformly charged thin spherical shell is constant and equal to the potential at its surface, given by the relation $V = \frac{kq}{R}$. Since the electric field inside a hollow conductor is zero, the potential remains uniform throughout the interior volume without changing from point to point. Because both points C and P are located in this interior region where the potential is invariant, their respective electric potentials are equal to each other. Subtracting these identical potential values results in a potential difference of zero.

The capacitance of a system is an intrinsic geometric property that depends entirely on the physical dimensions of the conductors, such as plate area and separation distance, along with the permittivity of the dielectric medium between them. Because capacitance is defined as the ratio of charge to potential difference, any change in charge results in a proportional change in potential, leaving their ratio constant. Consequently, capacitance remains completely independent of both the instantaneous charge stored and the applied potential difference across the plates.

Gauss law states that the total electric flux passing through any closed hypothetical surface depends exclusively on the net electrical charge contained within that volume. Geometrical characteristics such as the surface area, shape, and overall volume of the enclosing boundary do not affect the total flux lines emerging from it. Therefore, the net flux is determined solely by the enclosed charge divided by the permittivity of free space.

Gauss's law relates the surface integral of an electric field to the total electric flux passing through a closed surface. When this surface integral equals zero, it indicates that the net electric flux through the entire closed boundary is zero. Physically, this means that every electric field line entering the enclosed volume must also exit it, resulting in a balanced flow of incoming and outgoing field lines.

The total electric potential at any point in space is determined by the algebraic sum of the scalar potentials contributed by individual charges, following the core relation $V = \frac{K q}{r}$. For the given charge configuration, the net potential at point $P$ is calculated by subtracting the potential due to the more distant charge from that of the closer charge, expressed as $V_{\text{net}} = \frac{K q}{2 \times 10^{-2}} - \frac{K q}{8 \times 10^{-2}}$. Rearranging and factoring this expression leads to $V_{\text{net}} = \frac{K q \times 10^2}{2} \left(1 - \frac{1}{4}\right)$, which directly simplifies to the required result of $\frac{3}{8} q K$ scaled by 10^2 V.

The total electric flux emerging from a closed cubic surface enclosing a central point charge is distributed symmetrically across all six identical faces of the cube. According to Gauss law, the total flux through the entire enclosure is expressed as $\Phi = \frac{q}{\epsilon_0}$, where $q$ incorporates the charge magnitude in microcoulombs. Dividing this total flux by the six faces of التفاصيل and accounting for the microcoulomb multiplier yields the flux passing through any single face as $\frac{Q}{6\epsilon_0} \times 10^{-6}$.

The electric potential on the surface of a charged conducting sphere is given by the formula $V = \frac{KQ}{R}$, while the electric field at an external point at distance $r$ is expressed as $E = \frac{KQ}{r^2}$. By expressing the product $KQ$ as $VR$ from the potential equation, this term can be substituted directly into the electric field formula. This yields $E = \frac{VR}{r^2}$, providing the required relationship between the field and surface potential.