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QUESTION: 1

The electric flux density and electric field intensity have which of the following relation?

Solution:

Answer: a

Explanation: The electric flux density is directly proportional to electric field intensity. The proportionality constant is permittivity. D=ε E. It is clear that both are in linear relationship.

QUESTION: 2

The electric field intensity is the negative gradient of the electric potential. State True/False

Solution:

Answer: a

Explanation: V = -∫E.dl is the integral form. On differentiating both sides, we get E = -Grad (V). Thus the electric field intensity is the negative gradient of the electric potential.

QUESTION: 3

Find the electric potential for an electric field 3units at a distance of 2m.

Solution:

Answer: c

Explanation: The electric field intensity is the ratio of electric potential to the distance. E = V/d. To get V = E X d = 3 X 2 = 6units.

QUESTION: 4

Find the potential at a point (4, 3, -6) for the function V = 2x^{2}y + 5z.

Solution:

Answer: b

Explanation: The electric potential for the function V = 2x^{2}y + 5z at the point (4, 3, -6) is given by V = 2(4)2(3) + 5(-6) = 96-30 = 66 units.

QUESTION: 5

Find the electric flux density surrounding a material with field intensity of 2xyz placed in transformer oil ( εr = 2.2) at the point P(1,2,3) is

(in 10^{-10} units)

Solution:

Answer: c

Explanation: D = εE, where ε = εo εr. The flux density is given by,

D = 8.854 X 10^{-12} X 2.2 X 2(1)(2)(3) = 2.33 X 10^{-10} units.

QUESTION: 6

If potential V = 20/(x^{2} + y^{2}). The electric field intensity for V is 40(x i + y j)/(x^{2} + y^{2})^{2}. State True/False.

Solution:

Answer: a

Explanation: E = -Grad (V) = -Grad(20/(x^{2} + y^{2})) = -(-40x i /(x^{2} + y^{2})^{2} – 40(y j)/(x^{2} + y^{2})^{2}) = 40(x i + y j)/(x^{2} + y^{2})^{2}. Thus the statement is true.

QUESTION: 7

Find the potential of the function V = 60cos θ/r at the point P(3, 60, 25).

Solution:

Answer: b

Explanation: Given V = 60cos θ/r. For r = 3m and θ = 60, we get V = 60cos 60/3 = 20cos 60 = 10 units.

QUESTION: 8

Find the work done moving a charge 2C having potential V = 24volts is

Solution:

Answer: d

Explanation: The work done is the product of charge and potential.

W = Q X V = 2 X 24 = 48 units.

QUESTION: 9

If the potential is given by, V = 10sin θ cosφ/r, find the density at the point P(2, π/2, 0) (in 10^{-12} units)

Solution:

Answer: b

Explanation: Since V is given find out E.E = -Grad(V) = – Grad(10sin θ cosφ/r). From E, we can easily compute D. D = εE = 8.854 X 10^{-12} X 5/2 = 22.13 units.

QUESTION: 10

If V = 2x^{2}y + 20z – 4/(x^{2} + y^{2}), find the density at A(6, -2.5, 3) in nC/m^{2}

Solution:

Answer: a

Explanation: Find E from V, E = -Grad (V). We get E at A(6,-2.5,3) as 59.97i – 71.98j -20k. Thus D = εE = 8.854 X 10^{-12} X

(59.97i – 71.98j -20k) = (0.531i – 0.6373j – 0.177k) nC/m^{2}.

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