Electrostatics – NEET UG PhysicsPractice Questions & PYQs

The equilibrium of a suspended charged pith ball is governed by the balance between electrostatic repulsion, string tension, and gravitational force, leading to the fundamental relation that the cube of the separation distance is directly proportional to the vertical height of the suspension. For each pith ball, the horizontal electrostatic repulsion balances the horizontal component of the tension, while the vertical component of the tension balances the downward gravitational force, expressed as $\tan\theta = \frac{F_e}{mg}$. Substituting the electrostatic force formula and relating the geometric angle to the separation and vertical height yields the cubic proportionality $r^3 \propto y$. When the strings are rigidly clamped at half the original height, the new vertical height becomes $y/2$, which upon substitution into the proportional relationship reduces the new equilibrium separation to $r/\sqrt[3]{2}$.

For a system of three collinear charges to remain in complete equilibrium, the net electrostatic force acting on any individual charge must be zero. Placing a charge at the midpoint between two identical charges requires balancing the repulsive force exerted by one outer charge on the other with the force from the central charge, leading to the condition $\frac{Q^2}{r^2} = -\frac{4Qq}{r^2}$. Solving this force balance equation demonstrates that the central charge must equal negative $Q$ divided by four.

The work done to rotate an electric dipole in a uniform electric field corresponds directly to the change in its electrostatic potential energy, defined by the relation $U = -pE(\cos\theta_2 - \cos\theta_1)$. Substituting the initial angle of zero degrees and the final angle of ninety degrees into this expression yields $U = -pE(\cos 90^{\circ} - \cos 0^{\circ})$. Since the cosine of ninety degrees is zero and the cosine of zero degrees is one, the equation simplifies to $-pE(0 - 1)$. This calculation results in a positive energy requirement of $pE$.

Electric potential always decreases as one moves along the direction of a uniform electric field. Because of this fundamental property, the potentials at these points follow the order $V_B > V_C > V_A$. Therefore, the electric potential achieves its maximum value at point B.

The electrostatic potential energy stored within a capacitor is determined by the expression $U = \frac{1}{2} C V^2$. For a parallel plate geometry, the capacitance is defined as $C = \frac{\varepsilon_0 A}{d}$ and the potential difference across the plates relates to the uniform electric field through $V = Ed$. Substituting these geometric and field relationships into the energy formula yields $U = \frac{1}{2} \left(\frac{\varepsilon_0 A}{d}\right) (Ed)^2$, which simplifies directly to $\frac{1}{2} \varepsilon_0 E^2 A d$.

Electric potential is a scalar quantity, meaning the total potential at the center ofзел the square is the algebraic sum of the potentials due to each individual charge, expressed by $V = \frac{1}{4\pi\varepsilon_0} \sum \frac{q_i}{r}$. Since the distance from each corner to the center is identical, the denominator and constants factor out completely. Equating the sum of the charges to zero yields $-Q - q + 2q + 2Q = 0$, which directly reduces to the relation $Q = -q$.

Gauss's law establishes that the total electric flux emerging from a closed surface depends exclusively on the charge enclosed within it. A charge positioned precisely at the corner of a cube is shared among eight adjacent corners of a larger symmetric lattice formed by identical cubes. Consequently, only an eighth of the total flux generated by the charge passes through the specific faces of the target cube, resulting in a net flux of $\phi = \frac{q}{8 \varepsilon_0}$.

When an electric dipole is placed in a uniform electric field, it experiences a torque due to the cross product of the dipole moment vector and the electric field vector, along with a potential energy determined by their dot product. Simultaneously, the electrostatic potential energy of the system is defined as the negative dot product of these two vectors, referenced to zero at a perpendicular orientation. Evaluating these standard vector relationships directly produces a torque expression of $pE\sin\theta$ and a potential energy expression of $-pE\cos\theta$.

Connecting two charged conductors with a wire forces charge to flow until both spheres attain an identical electric potential. Expressing this equality of surface potentials gives $\frac{k Q_1}{r_1} = \frac{k Q_2}{r_2}$, which indicates that the charges are directly proportional to the radii of the spheres so that $Q_1 = 3Q_2$. Applying the conservation of total charge, where the initial sum equals $4 \times 10^{-2}$ C, allows us to solve for the individual charges to find that the final charge on the bigger sphere is $3 \times 10^{-2}$ C.

Gauss's law establishes that total electric flux is determined solely by the net charge enclosed within a surface divided by the permittivity of free space, as represented by $\Phi = \frac{Q}{\varepsilon_0}$. Expanding the radius of the spherical Gaussian surface alters neither the magnitude of deduced enclosed charge nor the fundamental nature of the electric field lines passing through it. Consequently, doubling the radius leaves the outward electric flux completely unchanged.

The electrostatic potential energy stored in a capacitor is given by the relation $U = \frac{1}{2} C V^{2}$, where $C$ is the capacitance and $V$ is the potential difference. For a parallel plate configuration, the capacitance is expressed as $C = \frac{\varepsilon_{0} A}{d}$ and the potential difference relates to the uniform electric field through $V = E d$. Substituting these expressions into the energy formula yields $U = \frac{1}{2} \left(\frac{\varepsilon_{0} A}{d}\right) (E d)^{2}$, which simplifies directly to $\frac{1}{2} \varepsilon_{0} E^{2} A d$ upon expansion.

The total electric potential at any location is determined by the principle of scalar superposition, combining the individual potentials from each source through the relation $V = \sum \frac{1}{4\pi\varepsilon_0}\frac{q_i}{r_i}$. Considering a square with side lengths of $2L$, the midpoint situated between the two positive charges lies at a distance of $L$ from each of those respective sources.

The remaining two negative charges reside at the opposing corners, establishing a distance from the target midpoint calculated via the Pythagorean theorem as $\sqrt{(2L)^2 + L^2} = L\sqrt{5}$. Combining all four contributions algebraically results in $\frac{1}{4\pi\varepsilon_0}\left(\frac{q}{L} + \frac{q}{L} - \frac{q}{L\sqrt{5}} - \frac{q}{L\sqrt{5}}\right)$. Factoring this combined expression yields the final value of $\frac{1}{4\pi\varepsilon_0}\frac{2q}{L}\left(1 - \frac{1}{\sqrt{5}}\right)$.

The work done in moving a test charge between two points in an electrostatic field is determined by the potential difference between them, given by the relation $W = Q(V_E - V_D)$. Evaluating this requires calculating the exact electric potentials generated by the three charges at both locations.

Since the triangle is isosceles with $AC = BC = 2a$ and the points D and E are located at the midpoints of their respective sides, the segments $BD$, $DC$, $AE$, and $EC$ each equal $a$. Using the geometric properties of the triangle, the remaining distances from D to the opposite vertex and from E to the opposite vertex are both found to be $a\sqrt{3}$ due to the inherent symmetry of the configuration. Summing the individual potential contributions from all three charges yields an identical electric potential value for both points, meaning $V_D = V_E$. Consequently, substituting these equal potentials into the work equation results in a net work done of zero.

Electrostatic equilibrium dictates that excess charge on a conducting spherical shell resides entirely on its outer surface, leaving no net charge enclosed within the hollow interior. By applying Gauss's law to any concentric Gaussian surface drawn inside the shell at a radius less than the outer radius, the enclosed charge evaluates to zero. As a result, the electric field at any point situated inside a charged conducting sphere, including at a distance of half its radius, remains completely zero regardless of the external field intensity.

Electric flux is defined by the relation $\Phi = \vec{E} \cdot \vec{A} = EA\cos\phi$, where $\phi$ represents the angle between the electric field vector and the normal area vector. Because the square surface and the uniform electric field both lie entirely within the plane of the paper, the area vector is directed perpendicular to this plane. The angle between the electric field and the area vector is therefore 90 degrees, which yields zero electric flux.

The electric field vector is determined from the negative spatial gradient of the electric potential, written as $\vec{E} = -\nabla V$. Differentiating the potential function $V = 4x^2$ with respect to $x$ produces $-8x$, whereas the partial derivatives with respect to $y$ and $z$ vanish. Substituting the coordinate point (1, 0, 2) into this derivative yields an electric field of $-8\hat{i}$ volt/meter, confirming a magnitude of 8 directed along the negative X-axis.

The total capacitance of $n_1$ identical capacitors connected in series is given by $C_s = \frac{C_1}{n_1}$, and the energy stored when charged to a potential difference of $4V$ is expressed as $U_s = \frac{1}{2} C_s (4V)^2$. Conversely, a parallel combination of $n_2$ capacitors of value $C_2$ yields an equivalent capacitance of $C_p = n_2 C_2$, storing an energy of $U_p = \frac{1}{2} C_p V^2$ when charged to $V$. Equating the two stored energy expressions and substituting the respective capacitances allows the equation to be solved for $C_2$, directly leading to $\frac{16C_1}{n_1n_2}$.

The electric field between parallel metal plates in a vacuum is given by $E = \frac{\sigma}{\varepsilon_0}$. When an insulating dielectric medium such as kerosene oil with a dielectric constant $K > 1$ is introduced between the plates, bound charges are induced that create an opposing internal field. Consequently, the net electric field reduces to $E' = \frac{\sigma}{\varepsilon_0 K}$, causing the overall field strength to decrease.

Electrostatic repulsion between two point charges is governed by Coulomb's law, expressed as $F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{d^2}$ for two identical charges $q$ separated by a distance $d$. Rearranging this equation to solve for the charge magnitude gives $q = \sqrt{4\pi\varepsilon_0 F d^2}$. Since charge is quantized as $n$ multiples of the elementary electronic charge $e$, the number of missing electrons is obtained by setting $q = ne$, resulting in $n = \frac{q}{e}$ which evaluates to $\sqrt{\frac{4\pi\varepsilon_0 F d^2}{e^2}}$.

The electric potential at any point due to a system of concentric spherical shells is the sum of potentials contributed by each individual shell, where points inside a shell experience a potential equal to that at its surface. Writing out the potential equations for each shell using their respective surface charge densities and radii allows us to evaluate their relative magnitudes. Given the condition $c = a + b$, we can test proportional values such as $a = a$, $b = 2a$, and $c = 3a$ to simplify the expressions. Substituting these values into the potential formulas demonstrates that the potentials for the innermost and outermost shells turn out to be equal, while the middle shell maintains a distinct potential value.