Electrostatics Interview Questions

Q1. State Gauss's law for electrostatics and explain its application to a point charge.

Gauss's law states that the total electric flux through any closed surface equals the enclosed free charge divided by permittivity: ∮D·dA = Qenc. For a point charge Q = 1 nC, applying Gauss's law to a spherical surface of radius 10 cm gives E = Q/(4πε0r²) = 899 V/m, which matches Coulomb's law derived directly. Gauss's law is the integral form of ∇·D = ρv and is most powerful when the charge distribution has spherical, cylindrical, or planar symmetry.

Follow-up: How does Gauss's law apply to finding the electric field inside and outside a uniformly charged sphere?

Q2. What is electric flux density D and how does it differ from electric field E?

Electric flux density D = εE = ε0εrE is the displacement field that accounts for bound charges in a dielectric, while E is the total force-per-unit-charge field that depends on both free and bound charges. In a parallel-plate capacitor with SiO2 dielectric (εr = 3.9), D at the plate surface equals the free surface charge density σ = Q/A regardless of the dielectric, while E = D/ε0εr is reduced by the factor εr compared to air. D is useful because its normal component is continuous across boundaries between dielectrics if there is no free surface charge.

Follow-up: What happens to D and E at the boundary between two different dielectric materials?

Q3. What is the capacitance of a parallel plate capacitor and what factors affect it?

Capacitance C = ε0εrA/d, where A is plate area, d is separation, and εr is the relative permittivity of the dielectric between the plates. A 1 cm × 1 cm capacitor with 100 nm SiO2 dielectric (εr = 3.9) has C = 8.854×10⁻¹² × 3.9 × 10⁻⁴/10⁻⁷ = 34.6 pF, which is the order of magnitude for gate capacitance in MOSFET devices. Reducing d or increasing εr both increase capacitance, which is why high-k dielectrics like HfO2 are used in modern CMOS nodes below 45 nm.

Follow-up: What is the maximum electric field a dielectric can withstand before breakdown, and what happens at breakdown?

Q4. What is the boundary condition for the electric field at a conductor surface?

At a perfect conductor surface, the tangential component of E is zero (Et = 0) and the normal component is E = σs/ε0, where σs is the surface charge density, because no electric field can exist inside a conductor in electrostatic equilibrium. The surface of a 10 kV electrode with charge density 0.1 μC/m² has E = 0.1×10⁻⁶/8.854×10⁻¹² = 11,295 V/m pointing normal to the surface. This is why conductor surfaces are equipotential surfaces and why sharp edges concentrate electric field causing corona discharge.

Follow-up: Why do sharp points on high-voltage conductors cause corona discharge at lower voltages than smooth surfaces?

Q5. What is Poisson's equation and when does it reduce to Laplace's equation?

Poisson's equation ∇²V = −ρv/ε describes the electric potential V in a region with volume charge density ρv; when ρv = 0 it reduces to Laplace's equation ∇²V = 0. In the channel region of a MOSFET with doping concentration Na, Poisson's equation ∇²ψ = qNa/ε gives the potential profile determining the depletion width. Laplace's equation is solved for potential between the plates of a capacitor where no free charge exists in the gap.

Follow-up: What is the method of images and which boundary value problems does it simplify?

Q6. What is electric potential and how is it related to electric field?

Electric potential V at a point is the work done per unit positive charge in bringing a test charge from infinity to that point, and E = −∇V, meaning E points in the direction of decreasing potential. For a point charge Q = 5 nC, the potential at 10 cm = Q/(4πε0r) = 449.4 V, and the electric field 449.4/0.1 = 4494 V/m points radially outward. The negative gradient relationship means potential and field lines are always perpendicular, which is why equipotential surfaces are perpendicular to field lines.

Follow-up: Why is it more convenient to work with scalar potential V rather than vector field E in many problems?

Q7. What is dielectric polarization and what causes it?

Dielectric polarization P is the electric dipole moment per unit volume created when an external electric field displaces bound charges in a dielectric material, described by P = ε0χeE = D − ε0E. In a BaTiO3 ceramic (εr ≈ 1700), an applied field of 1 MV/m induces P = ε0 × 1699 × 10⁶ = 15 mC/m², creating strong bound surface charges that dominate the capacitance. Polarization is the microscopic origin of the dielectric constant and explains why capacitors with high-k materials store more energy than air capacitors.

Follow-up: What is the difference between electronic, ionic, and orientational polarization mechanisms?

Q8. What is the energy stored in a capacitor and in an electric field?

Energy stored in a capacitor is W = ½CV² = Q²/(2C) = ½QV, and the energy density stored in an electric field is w = ½εE² J/m³. A 100 μF capacitor charged to 400 V stores W = ½ × 100×10⁻⁶ × 160,000 = 8 J, which is the energy released in the flash of a camera. The volumetric energy density formula w = ½εE² explains why increasing breakdown field strength is the key metric for improving energy storage density in capacitor films.

Follow-up: What happens to energy stored in a capacitor when the dielectric is inserted with the capacitor connected to a voltage source versus disconnected?

Q9. What is the method of images in electrostatics?

The method of images replaces a grounded conductor with an image charge of opposite sign placed symmetrically behind the conductor surface, simplifying the boundary value problem by creating the same boundary condition without the conductor present. A charge +Q at height h above a grounded infinite plane is equivalent to +Q and an image −Q at depth h below the plane, giving E and V in the upper half-space directly. This method is widely used in analyzing electrostatic discharge (ESD) events near PCB ground planes and in transmission line calculations.

Follow-up: Can the method of images be applied to a charge near a spherical conductor?

Q10. What is capacitance of a coaxial cable and how is it derived?

Using Gauss's law, E between the inner conductor (radius a) and outer conductor (radius b) of a coaxial cable is E = ρL/(2πεr) in the radial direction, and integrating gives V = ρL·ln(b/a)/(2πε); capacitance per unit length is C = 2πε/ln(b/a). An RG-58 coaxial cable with a = 0.45 mm, b = 1.5 mm, and εr = 2.3 has C ≈ 2π × 2.3 × 8.854×10⁻¹²/ln(3.33) ≈ 107 pF/m, which matches its datasheet specification. This capacitance is critical in high-speed signal integrity as it determines the characteristic impedance along with inductance per unit length.

Follow-up: How does the capacitance per unit length of a coaxial cable relate to its characteristic impedance?

Q11. What is the electric field inside a hollow charged conductor?

The electric field inside a hollow conductor is exactly zero in electrostatic equilibrium, regardless of the external charge distribution or external fields, because free electrons redistribute themselves on the outer surface to cancel any internal field. This is the Faraday cage effect, exploited in shielded RF enclosures and screened rooms where the conductor shell is aluminum or copper of sufficient thickness. Even with 10,000 V applied to the outer shell, a person or sensitive instrument inside a Faraday cage experiences zero electric field.

Follow-up: Does a Faraday cage also shield against magnetic fields, and why or why not?

Q12. What is Coulomb's law and what are its limitations?

Coulomb's law states that the force between two point charges is F = Q1Q2/(4πε0r²) along the line joining them, attractive for unlike charges and repulsive for like charges. Two 1 μC charges separated by 1 cm experience F = (10⁻⁶)²/(4π × 8.854×10⁻¹² × 10⁻⁴) = 89.9 N, which is surprisingly large given the charge magnitudes. Coulomb's law applies only to static charges; moving charges create magnetic fields and radiation effects that Coulomb's law cannot describe, requiring the full Maxwell framework.

Follow-up: How does Coulomb's law change in a dielectric medium compared to free space?

Q13. What is the superposition principle in electrostatics?

The superposition principle states that the electric field or potential due to a collection of charges equals the vector sum of the fields (or scalar sum of potentials) due to each individual charge alone, because Maxwell's equations are linear. For an array of 4 point charges at the corners of a square of side 10 cm, the field at the center is found by computing E from each charge separately and adding all four vectors. This principle underlies the validity of circuit analysis with multiple sources and is the foundation for method of moments numerical solvers in antenna and ESD analysis.

Follow-up: Does the superposition principle hold in nonlinear dielectric materials?

Q14. What is the significance of the electric dipole and dipole moment?

An electric dipole consists of equal and opposite charges ±Q separated by a small distance d, with dipole moment p = Qd pointing from −Q to +Q, and it is the simplest non-trivial electrostatic source with field falling as 1/r³ rather than 1/r². A water molecule has a permanent dipole moment of 6.17×10⁻³⁰ C·m (1.85 Debye), which is responsible for water's high dielectric constant of 80 and its effectiveness as a polar solvent. Dipole radiation is the dominant mechanism of electromagnetic emission from small antennas and switching circuits.

Follow-up: What is a quadrupole and in what situation is its field important?

Q15. What happens to charge distribution on a conductor with a sharp corner or point?

Surface charge density is highest at points and sharp corners of a conductor because the electric potential must be constant across the surface, requiring the field (proportional to surface charge density) to be highest where the radius of curvature is smallest. The tip of a lightning rod at 100 kV with tip radius 1 mm can have E exceeding 10 MV/m, initiating ionization and corona discharge that bleeds charge from clouds. This concentration effect is why high-voltage equipment uses large-radius toroids and smooth profiles, and why ESD from a fingertip (sharp point) can destroy a MOSFET gate oxide.

Follow-up: What is the breakdown strength of air and at what voltage does a 1 mm air gap break down?