Test: Electric Field Intensity - Electrical Engineering (EE) MCQ

Answer: c
Explanation: The electric field intensity is the force per unit charge on a test charge, i.e, q1 = 1C. E = F/Q = Q/(4∏εr2).

Answer: c
Explanation: Force is the product of charge and electric field.
F = q X E = 2 X 1 = 2 N.

Answer: b
Explanation: F = q1q2/(4∏εor2) = -2 X 9/(10^-9 X 12) = -18 X 10⁹
E = F/q = 18 X 10⁹/2 = 9 X 10⁹.

Answer: c
Explanation: E = Q/ (4∏εor2)
= (2 X 10^-6)/(4∏ X εo X 0.2²) = 450,000 V/m

Answer: a
Explanation: E = Q/ (4∏εor²)
Q = (4000 X 0.3²)/ (9 X 10⁹) = 4 X 10^-8 C.

Answer: a
Explanation: If a test charge +q is situated at a distance r from Q, the test charge will experience a repulsive force directed radially outward from Q. Since electric field is inversely proportional to distance, thus the statement is true.

Answer: a
Explanation: The electric field intensity of an infinitely long conductor is given by, E = λ/(4πεh).(sin α2 – sin α1)i + (cos α2 + cos α1)j
For an infinitely long conductor, α = 0. E = λ/(4πεh).(cos 0 + cos 0) = λ/(2πεh).aN.

Answer: d
Explanation: E = σ/2ε.(1- cos α), where α = h/(√(h²+a²))
Here, h is the distance of the sheet from point P and a is the radius of the sheet. For infinite sheet, α = 90. Thus E = σ/2ε.

Answer: c
Explanation: E = Q/ (4∏εor²)
When distance d is infinity, the electric field will be zero, E= 0.

Answer: c
Explanation: In an electromagnetic wave, the electric field and magnetic field will be perpendicular to each other. Both of these fields will be perpendicular to the wave propagation.