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QUESTION: 1

The electric field intensity is defined as

Solution:

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).

QUESTION: 2

Find the force on a charge 2C in a field 1V/m.

Solution:

Answer: c

Explanation: Force is the product of charge and electric field.

F = q X E = 2 X 1 = 2 N.

QUESTION: 3

Find the electric field intensity of two charges 2C and -1C separated by a distance 1m in air.

Solution:

Answer: b

Explanation: F = q1q2/(4∏εor2) = -2 X 9/(10^{-9} X 12) = -18 X 10^{9}

E = F/q = 18 X 10^{9}/2 = 9 X 10^{9}.

QUESTION: 4

What is the electric field intensity at a distance of 20cm from a charge 2 X 10^{-6} C in vacuum?

Solution:

Answer: c

Explanation: E = Q/ (4∏εor2)

= (2 X 10^{-6})/(4∏ X εo X 0.2^{2}) = 450,000 V/m

QUESTION: 5

Determine the charge that produces an electric field strength of 40 V/cm at a distance of 30cm in vacuum(in 10-8C)

Solution:

Answer: a

Explanation: E = Q/ (4∏εor^{2})

Q = (4000 X 0.3^{2})/ (9 X 10^{9}) = 4 X 10^{-8} C.

QUESTION: 6

The field intensity of a charge defines the impact of the charge on a test charge placed at a distance. It is maximum at d = 0cm and minimises as d increases. State True/False

Solution:

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.

QUESTION: 7

Electric field of an infinitely long conductor of charge density λ, is given by E = λ/(2πεh).aN. State True/False.

Solution:

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.

QUESTION: 8

Electric field intensity due to infinite sheet of charge σ is

Solution:

Answer: d

Explanation: E = σ/2ε.(1- cos α), where α = h/(√(h^{2}+a^{2}))

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.

QUESTION: 9

For a test charge placed at infinity, the electric field will be

Solution:

Answer: c

Explanation: E = Q/ (4∏εor^{2})

When distance d is infinity, the electric field will be zero, E= 0.

QUESTION: 10

In electromagnetic waves, the electric field will be perpendicular to which of the following?

Solution:

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.

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