Electrostatics – NEET UG PhysicsPractice Questions & PYQs

An infinite line of charge generates a radial electric field whose magnitude at a perpendicular distance $r$ is expressed as $E = \frac{\lambda}{2\pi \varepsilon_0 r}$. At the exact midpoint between two parallel line charges carrying equal and opposite densities of $+\lambda$ and $-\lambda$ separated by a distance of $2R$, the distance to each line is $R$. The electric field vectors from both line sources point in the identical direction-away from the positive charge and toward the negative charge. Summing these two equal contributions yields a net electric field magnitude of $\frac{\lambda}{\pi \varepsilon_0 R}$.

When two conducting spheres are brought into physical contact, charge redistributes until both spheres attain a common electrostatic potential. The total initial charge on the system is the sum of the individual charges calculated from the initial surface charge density and respective surface areas, giving a combined charge of $20\pi R^2 \sigma$. Equating the potentials of the spheres after contact dictates that their charges are proportional to their radii, leading to individual charges of $\frac{20}{3}\pi R^2 \sigma$ and $\frac{40}{3}\pi R^2 \sigma$. Dividing these newly redistributed charges by their respective surface areas yields the updated surface charge densities of $\frac{5}{3}\sigma$ and $\frac{5}{6}\sigma$.

Gauss's law dictates that any excess charge placed on a hollow conducting sphere resides entirely on its outer surface, leaving no enclosed charge within the interior cavity. Consequently, the electric field is strictly zero everywhere inside the sphere where $r < R$. Beyond the surface, the shell behaves electrostatically like a point charge concentrated at its geometric center, causing the electric field to fall off as $E \propto \frac{1}{r^2}$ for $r > R$.

When two capacitors are connected together, charge redistribution always results in a dissipation of electrostatic energy because of heat and radiation losses in the connecting wires. Initially, capacitor $C_1$ connects to the battery of emf $V$, storing an initial energy of $U = \frac{1}{2} C V^2$. When the switch shifts to connect with the identical uncharged capacitor $C_2$, the charge redistributes between them, and the energy loss evaluates to $\Delta U = \frac{C_1 C_2}{2(C_1 + C_2)} (V - 0)^2$, which simplifies to $\frac{1}{4} C V^2$. Comparing this energy loss directly to the initial stored energy yields a fraction of one-half, representing a 50 percent loss of the system's total energy.

The electrostatic force between the plates of a parallel plate capacitor is derived from the interaction between the charge on one plate and the electric field produced by the other plate. Since the electric field created by an infinite plane sheet of charge is given by $E = \frac{Q}{2A\varepsilon_0}$ and is independent of the distance from the sheet, the resulting force $F = \frac{Q^2}{2A\varepsilon_0}$ remains constant regardless of how far apart the plates are moved. Consequently, as long as the charge on an isolated capacitor is held constant, the attractive force between its plates does not depend on the separation distance.

A charged particle released from rest in a uniform electric field experiences a constant acceleration determined by the electrostatic force relative to its mass, leading to the kinematic time relation $t = \sqrt{\frac{2hm}{qE}}$. Because both an electron and a proton carry identical elemental charge magnitudes and traverse the same vertical displacement under the same field strength, their fall time depends exclusively on the square root of their respective masses. Since the mass of an electron is significantly smaller than that of a proton, the electron accelerates much faster and completes the fall in a smaller time.

The acceleration of the toy car during the first second is found using the kinematic formula $a = \frac{v - u}{t}$, yielding $6 \text{ m/s}^2$. During the initial interval from $0$ to $1 \text{ second}$, the car travels a forward displacement of $3 \text{ meters}$ using the relation $S = ut + \frac{1}{2}at^2$. When the electric field reverses direction at $t = 1 \text{ second}$, the car decelerates to a stop at $t = 2 \text{ seconds}$, covering another $3 \text{ meters}$ forward, and then travels backward by $3 \text{ meters}$ during the final interval from $t = 2$ to $3 \text{ seconds}$. Summing these segments gives a net displacement of $3 \text{ meters}$ and a total distance traveled of $9 \text{ meters}$. Dividing the net displacement by the total time of $3 \text{ seconds}$ gives an average velocity of $1 \text{ m/s}$, while dividing the total distance by the total time yields an average speed of $3 \text{ m/s}$.

The initial electrostatic energy stored in a charged capacitor is given by the relation $U_i = \frac{1}{2}CV^2$. When an identical uncharged capacitor is connected in parallel, the total capacitance doubles to $2C$, while the charge redistributes equally, causing the shared potential difference across each capacitor to drop to half of its original value. Substituting these new values into the energy formula gives a final electrostatic energy of $U_f = \frac{1}{2}(2C)\left(\frac{V}{2}\right)^2$, which simplifies to half of the initial energy. Thus, sharing the charge with an identical capacitor halves the total stored energy of the system.

The equilibrium condition where gravitational attraction exactly cancels out net electrostatic repulsion between two neutral atoms is expressed by the force balance equation $\frac{G m^2}{r^2} = \frac{k (\Delta e)^2}{r^2}$. Rearranging this expression to solve for the charge imbalance gives $\Delta e = m \sqrt{\frac{G}{k}}$. Substituting the mass of a hydrogen atom along with the gravitational and electrostatic constants calculates a charge difference on the order of $10^{-37} \text{ C}$.

The work done in moving a charge between two points is determined by the formula $W = q\Delta V$, where $\Delta V$ represents the potential difference between the initial and final equipotential surfaces. Since both the charge and the potential difference between the given points remain constant across all four diagrams, the energy required is identical in every case. Consequently, the work done is the same across all scenarios regardless of the specific configuration shown.

When multiple dielectric media partition the interior of a parallel-plate capacitor, the system resolves into a network of parallel and series capacitor combinations. The first three dielectric slabs divide the total cross-sectional area and operate as three capacitors connected in parallel, combining into an equivalent capacitance given by $C_A = \frac{2A\epsilon_0}{3d}(k_1 + k_2 + k_3)$.

This parallel group is arranged in series with the fourth dielectric layer, which possesses a capacitance of $C_B = \frac{A\epsilon_0 k_4}{d}$. Utilizing the standard reciprocal rule for series capacitance alongside the expression for a uniformly filled capacitor $C = \frac{k\epsilon_0 A}{d}$, we substitute these values to write $\frac{1}{C} = \frac{1}{C_A} + \frac{1}{C_B}$. Rearranging terms yields the final expression relating the unified dielectric constant to the individual values.

The mechanical equilibrium of each suspended sphere requires balancing the electrostatic repulsion, string tension, and gravitational force through the relations $T \sin\theta = \frac{kq^2}{x^2}$ and $T \cos\theta = mg$. Dividing these equations gives $\tan\theta = \frac{kq^2}{mgx^2}$, and for small angles, $\tan\theta$ can be approximated as $\frac{x}{2l}$. Equating this geometric approximation to the force ratio yields a proportional relationship where charge scales as $q \propto x^{3/2}$. Differentiating this relation with respect to time under the condition of a constant leak rate produces the velocity dependency $v = \frac{dx}{dt} \propto x^{-1/2}$.

The torque experienced by an electric dipole in a uniform electric field is governed by the relation $\tau = pE \sin \theta$, where $p$ is the dipole moment. Substituting the given values for torque, electric field, and angle into this expression allows us to isolate and calculate the dipole moment as $4 \times 10^{-5} \text{ C}\cdot\text{m}$. Since the dipole moment is defined as the product of the charge and the separation distance between the charges, dividing this dipole moment by the given length of $2 \text{ cm}$ yields a charge of $2 \times 10^{-3} \text{ C}$, which is equivalent to $2 \text{ mC}$.

When a charged capacitor is connected to an uncharged capacitor, charge redistribution takes place, resulting in a loss of electrostatic potential energy that is dissipated as heat in the connecting wires. The magnitude of this energy loss is evaluated using the redistribution formula
$\Delta E = \frac{C_1 C_2}{2(C_1 + C_2)}(V_1 - V_2)^2$
where the two interacting capacitances are $2$ microfarads and $8$ microfarads, subjected to initial potentials of $V$ and zero respectively.

The initial energy stored in the primary capacitor is given by $\frac{1}{2} C_1 V^2$. Substituting the respective capacitance values into the energy loss equation yields the exact amount of energy dissipated when the switch changes position. Expressing this dissipated energy as a fraction of the initial stored energy and multiplying by one hundred determines that the proportion of energy lost is $80\%$.

Gauss's law relates the total electric flux through a closed surface to the enclosed charge via the equation $\Phi = \oint \vec{E} \cdot d\vec{S} = \frac{q_{\text{in}}}{\epsilon_0}$. Evaluating this for a spherical surface of radius $a$ where the electric field is given by $E = Ar$, the magnitude of the field on the surface becomes $Aa$. Since the electric field and the normal vector to the surface are parallel everywhere, the surface integral simplifies to the product of the electric field magnitude and the total surface area of the sphere, $4\pi a^2$. Substituting these values yields a total flux of $4\pi A a^3$, which upon multiplication by $\epsilon_0$ gives the enclosed charge.

When a capacitor is charged by a source and subsequently disconnected, the isolated nature of the plates ensures that the total charge remains strictly constant. Introducing a dielectric material with constant $K$ into the gap increases the capacitance to $KC$ and polarizes the medium to reduce the internal electric field. Consequently, the stored energy and the potential difference between the plates both decrease by a factor of $K$. Therefore, asserting that the charge is not conserved contradicts the fundamental principle of charge conservation for an isolated system.

The electric field vector is derived from the negative spatial gradient of the electric potential using the differential operator $\vec{E} = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right)$. Taking partial derivatives of the given potential function $V(x, y, z) = 6xy - y + 2yz$ with respect to each coordinate yields $6y$ for the x-component, $6x - 1 + 2z$ for the y-component, and $2y$ for the z-component. Substituting the point coordinates (1, 1, 0) into these partial derivative expressions gives 6, 5, and 2 respectively. Combining these evaluated components with the negative gradient sign results in the final electric field vector $-\left(6\hat{i} + 5\hat{j} + 2\hat{k}\right)$.

The electric field inside any charged conductor is strictly zero because excess charges reside entirely on the outer surface, causing internal electric fields to cancel out completely. Since the electric field vanishes everywhere inside the volume, no work is required to move a test charge from the centre to the surface, meaning the potential at the centre is uniform and identical to the potential at the boundary. Therefore, the potential evaluates to $\frac{Q}{4\pi \varepsilon_0R}$ while the electric field is zero.

The electric field components are obtained by taking the negative partial derivatives of the potential function with respect to $x$, $y$, and $z$ respectively. Differentiating the given potential yields $E_x = -(6 - 8y)$, $E_y = -(-8x - 8 + 6z)$, and $E_z = -6y$. Evaluating these partial derivatives at the point $(1, 1, 1)$ gives $E_x = 2$, $E_y = 10$, and $E_z = -6$. The magnitude of the net electric field is calculated using $E = \sqrt{E_x^2 + E_y^2 + E_z^2}$, which results in $\sqrt{4 + 100 + 36} = 2\sqrt{35}$ volts per meter. Multiplying this electric field magnitude by a charge of 2 coulombs yields an electric force of $4\sqrt{35}$ newtons.

The electric field inside a parallel plate capacitor is defined by the formula $E = \frac{Q}{A\varepsilon_0}$ \in the absence of a medium, and modifies to $E = \frac{Q}{KA\varepsilon_0}$ when a dielectric is present. Because the dielectric constant is inversely proportional to the electric field strength, a larger dielectric value creates a weaker field between the plates. Given that $K_1 < K_2$, the electric field will maintain a higher magnitude in the first slab and drop to a lower, constant magnitude when transitioning into the second slab.