Electrostatics – NEET UG PhysicsPractice Questions & PYQs

When excess charge resides on a conductor in electrostatic equilibrium, charges rearrange themselves such that the internal electric field vanishes completely everywhere inside the material. Because the electric field represents the spatial rate of change of electric potential, a zero field implies that the potential has a zero gradient throughout the interior and surface. Consequently, the electric potential remains constant and uniform at every single point inside and across the conductor.

The electric potential energy of a central charge interacting with multiple surrounding charges is the sum of the individual pair-wise potential energies. The distance from the center of a cube to any of its eight corners is equal to half the body diagonal, which evaluates to $r = \frac{\sqrt{3} b}{2}$. Substituting this distance and the respective charges into the Coulomb potential energy formula gives the interaction energy with a single corner charge. Multiplying this individual energy by the eight identical corners yields the total electrostatic potential energy, simplifying to $U = \frac{-4 q^{2}}{\sqrt{3}\pi\varepsilon_{0} b}$.

Electrostatic energy density represents the energy stored per unit volume in the electric field between capacitor plates, which is fundamentally expressed as $\frac{1}{2} \varepsilon_0 E^2$. For a uniform field, the electric field strength is related to the potential difference and plate separation through the equation $E = \frac{V}{d}$. Substituting this field expression into the energy density formula yields $\frac{1}{2} \varepsilon_{0} \frac{V^{2}}{d^{2}}$.

According to Gauss's law, the total electric flux emanating through any closed surface enclosing a charge equals the enclosed charge divided by the permittivity of free space. Since a cube is a symmetrical enclosed surface with six identical faces, the total flux distributes equally across all of them. Dividing the total flux expression by the six faces of the cube, alongside the necessary unit conversion factor, gives $\frac{Q}{6 \varepsilon_{0}} \times 10^{-6}$ for each individual face.

When an electric dipole is placed in a uniform electric field, the equal and opposite forces acting on its charges produce a turning effect defined mathematically as a vector cross product. The magnitude of this rotational effect depends on the dipole moment magnitude, the electric field strength, and the sine of the angle between their directions. Vectorially, this relationship is expressed as $\overset{\rightarrow}{\tau} = \overset{\rightarrow}{p} \times \overset{\rightarrow}{E}$.

Gauss's law can be applied to corner positions by imagining a point charge as being shared symmetrically among eight identical cubes meeting at that single spatial vertex. The three faces of the cube that intersect directly at that corner contain the electric field lines, resulting in zero flux passing through them. Consequently, the net flux associated with the cube passes exclusively through the remaining faces, yielding a total electric flux of $\frac{Q}{8 \varepsilon_{0}}$ for the cube.

The capacitance of a parallel plate capacitor is directly proportional to the dielectric constant of the medium filling the space between its plates. The initial capacitance with oil is given by $C = \frac{K \varepsilon_{0} A}{d}$, which accounts for the dielectric constant value of $K = 2$. When the oil is removed and replaced by air, the capacitance is divided by this dielectric constant, resulting in $C' = \frac{C}{K} = \frac{C}{2}$.

The electric field due to an infinitesimal charge element $dq = \lambda a \, d\theta$ on a semicircular arc at its center is expressed as:
$dE = \frac{1}{4\pi\varepsilon_0} \frac{dq}{a^2} = \frac{\lambda \, d\theta}{4\pi\varepsilon_0 a}$

By symmetry, the horizontal components of the electric field cancel out, leaving only the vertical components to contribute to the net field. The total electric field is obtained by integrating the vertical component $dE \sin\theta$ from $0$ to $\pi$:
$E = \int_0^\pi \frac{\lambda}{4\pi\varepsilon_0 a} \sin\theta \, d\theta$

Evaluating this integral yields:
$E = \frac{\lambda}{4\pi\varepsilon_0 a} [-\cos\theta]_0^\pi = \frac{\lambda}{4\pi\varepsilon_0 a} (1 - (-1)) = \frac{\lambda}{2\pi\varepsilon_0 a}$

The initial electrostatic energy stored in a charged capacitor is given by the relation $U = \frac{1}{2} \frac{q^{2}}{C}$ in terms of its charge and capacitance. When the charging battery is disconnected and an identical uncharged capacitor is connected in parallel, the total charge $q$ redistributes equally, leaving each capacitor with a charge of half the original amount. Substituting this reduced charge back into the energy formula shows that the energy stored in each individual capacitor reduces to $\frac{U}{4}$.

The work done in moving a charge through a uniform electric field is defined by the relation $W = q E d \cos\theta$, where $\theta$ represents the angle between the field direction and the displacement vector. Rearranging this work-energy expression allows us to solve directly for the electric field magnitude by dividing the work done by the product of the charge, displacement, and the cosine of the given angle. Substituting the specified values of work, charge, distance, and angle into the rearranged formula gives the final field strength of $20$ N/C.