Physics

Electrostatics

299 Questions

Electrostatics deals with electric charges, fields, and potentials at rest. It is a crucial topic for physics sections in engineering and civil services competitive examinations. Review these questions to build a strong understanding of Coulomb law, electric dipoles, Gauss law, and electric flux.

Electric field and potentialElectric dipole momentGauss Law applicationsCoulomb force calculationsCharge distribution on spheresEquipotential surfaces

Electrostatics Questions

Multiple choice general knowledge science & technology

The primary law for electrostatics

  1. Charle' law

  2. Coulomb' Law

  3. Chandrasekhar limit

  4. Coanda effect
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Coulomb's Law (1785) is the fundamental law of electrostatics, quantifying force between electric charges: F = k(q₁q₂/r²)r̂, where k = 1/4πε₀. Charles' law is a gas law, Chandrasekhar limit concerns stellar astrophysics, and Coanda effect is fluid dynamics.

Multiple choice

Medium 1 has the electrical permittivity $\xi_1$= 1.5 $\xi_0$ farad/m and occupies the region to the left of x = 0 plane. Medium 2 has the electrical permittivity $\xi_2$ = 2.5 $\xi_0$ farad/m and occupies the region to the right of x = 0 plane. If E1 in medium 1 is E1 = (2ux -3uy + 1uz) volt/m, then E2 in medium 2 is

  1. (2.0ux - 7.5uy + 2.5uz) volt/m

  2. (2.0ux - 2.0uy + 0.6uz) volt/m

  3. (1.2ux - 3.0uy + 1.0uz) volt/m

  4. (1.2ux - 2.0uy + 0.6uz) volt/m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Multiple choice

A medium is divided into regions I and II about x = 0 plane, as shown in the figure below. An electromagnetic wave with electric field E1 = 4$\widehat a$x + 3$\widehat a$y + 5$\widehat a$z is incident normally on the interface form region-I. The electric field E2 in region-II at the interface is

  1. E2 = E1
  2. 4$\widehat a$x + 0.75$\widehat a$y + 1.2 5$\widehat a$z
  3. 3$\widehat a$x + 3$\widehat a$y + 5$\widehat a$z
  4. - 3$\widehat a$x + 3$\widehat a$y + 5$\widehat a$z
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Multiple choice

Two identical antennas are placed in the $\theta$= $\dfrac{\pi}{2}$ plane as shown in figure. The elements have equal amplitude excitation with 180° polarity difference, operating at wavelength $\lambda$. The correct value of the magnitude of the far-zone resultant electric field strength normalized with that of a single element, both computed for $\phi$= 0, is
  1. 2 cos $\left( \dfrac{2\pi s}{\lambda} \right)$
  2. 2 sin $\left( \dfrac{2\pi s}{\lambda} \right)$
  3. 2 cos $\left( \dfrac{\pi s}{\lambda} \right)$
  4. 2 sin $\left( \dfrac{\pi s}{\lambda} \right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Multiple choice

Which of the following is a vector?

  1. Electric charge

  2. Electric potential

  3. Intensity of the electric field

  4. Electrical resistance
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Electric field intensity is a vector quantity because it has both magnitude and direction - it points in the direction a positive test charge would move. Electric charge, potential, and resistance are all scalar quantities with magnitude but no direction.

Multiple choice

A semi-circular arc of radius r is charged uniformly and the charge per unit length is λ. The electric field at its centre is

  1. λ/2πє0r2

  2. λ/4πє0r2

  3. λ/3πє0r

  4. λ/2π є0r

  5. λ/21π є0r
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Electric intensity at centre O, due to small element dl of the charged ring dE = λ dl / 4πє0r2 = λ(a dӨ) / 4πє0r2.  As is clear from figure, horizontal components of dE will cancel out in pairs and vertical components will add. π π ... E = ∫dE sinӨ = ∫ λ / 4πє0r (sinӨ dӨ) 0 0 π = λ / 4πє0r [ - cos Ө] o dl dE E = λ / 4πє0r ( - cosπ + cos0) = λ / 2π є0r.