EduRev

Courses

Get the App Signup / Login

Welcome Back

Forgot Password?

New User? Sign Up

Create your account now

I accept the Terms & Conditions.

Already Have an Account? Login

Test: Electrostatics- 2

20 Questions MCQ Test Topicwise Question Bank for Electrical Engineering | Test: Electrostatics- 2

Description

This mock test of Test: Electrostatics- 2 for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 20 Multiple Choice Questions for Electrical Engineering (EE) Test: Electrostatics- 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Electrostatics- 2 quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Electrostatics- 2 exercise for a better result in the exam. You can find other Test: Electrostatics- 2 extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.

QUESTION: 1

Which one of the following pairs is NOT correctly matched?

Solution:

Option (c) is Faraday's law given by

QUESTION: 2

Match List-I (Law/Quantity) with List-ll (Mathematical expression) and select the correct answer using the codes given below the lists:
List-I
A. Gauss’s law
B. Ampere’s law
C. Faraday’s law
D. Poynting vector

Codes:
A B C D
(a) 1 2 4 3
(b) 3 5 2 1
(c) 1 5 2 3
(d) 3 2 4 1

A.
(a)
B.
(b)
C.
(c)
D.
(d)

Solution:

QUESTION: 3

What is the value of total electric flux coming out of a closed surface?

A.
Zero
B.
Equal to volume charge density
C.
Equal to the total charge enclosed by the surface
D.
Equal to the surface charge density

Solution:

From Gauss’s law, total electric flux from a closed surface is
.

QUESTION: 4

Which one of the following is the Poisson's equation for a linear and isotropic but in homogeneous medium?

Solution:

For homogeneous medium ∇² V = -ρ∈ and for non-homogeneous medium ∈ is a variable quantity due to which Poisson’s equation becomes

QUESTION: 5

Two concentric spherical shells carry equal and opposite uniformly distributed charges over their surfaces as shown in figure below.

Electric field on the surface of inner shell will be

A.
zero
B.
C.
D.

Solution:

QUESTION: 6

Electric field intensity due to line charge of infinite length is

Solution:

QUESTION: 7

Consider the following statements associated with equipotential surface:
1. Potential is same everywhere.
2. No current flows on this surface.
3. Work done in moving charge from one point to another is zero.
4. Potential is different everywhere.
Which of the above statements is/are not correct?

A.
1 and 3 only
B.
3 and 4 only
C.
4 only
D.
2 and 4 only

Solution:

Potential is same everywhere on an equipotential surface. Hence, statement 4 is not correct. Statements 1,2 and 3 are correct.

QUESTION: 8

A finite sheet 0 ≤ x ≤ 1, 0 ≤ 1 on the z = 0 plane has a charge density
ρ_s = xy (x² + y² + 25)^3/2 nC/m².
The total charge on the sheet would be

Solution:

The total charge on the sheet is

Since, , we now integrate w.r.t. x
(or change variables: x² = u so that xdx = du/2)

QUESTION: 9

Assertion (A): Gauss’s law is an alternative statement of Coulomb’s law.
Reason (R): Proper application of the divergence theorem to Coulomb’s law results in Gauss’s law.

A.
Both A and R are true and R is a correct explanation of A.
B.
Both A and R are true but R is not a correct explanation of A.
C.
A is true but R is false.
D.
A is false but R is true.

Solution:

QUESTION: 10

If the potential V = 10/r² sinθcosϕ then the work done in moving a 10 μC charge from point A(1, 30°, 120°) to B(4, 90°, 60°) would be equal to

A.
16.65 μJ
B.
28.125μJ
C.
25,25 μJ
D.
30.625 μJ

Solution:

Required work done,

QUESTION: 11

A wire of diameter 1 mm and conductivity 5 x 10⁷S/m has 10²⁹ free electrons /m³when an electric field of 10 mV/m is applied.
The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10^-19 C)

A.
500 kA/m²and 3.125 x 10^-5 m/s
B.
300 kA/m² and 2.225 x 10^-5m/s
C.
500 kA/m² and 2.225 x 10^-5m/s
D.
300 kA/m² and 3.125 x 10^-5 m/s

Solution:

The charge density of free electron is:
ρ_v = ne = (10²⁹) x (-1.6 x 10^-19)
= -1.6 x 10¹⁰ C/m²
Given,
E = 10 mV/m,
d = 1 mm,
σ = 5 x 10⁷ S/m,
n = 10²⁹/m³
Now, current density,

Also, drift velocity,

QUESTION: 12

The equivalent capacitance between the terminals A and B for the capacitor shown below would be (the area occupied by each dielectric is 30 cm²)

A.
26.53 pF
B.
28.68 pF
C.
25.46 pF
D.
22.22 pF

Solution:

Given,
d = 5 mm, d/2 = 2.5 mm,
∈_f1 = 4, ∈_f2 = 6,
A/2 = 15 cm²
For the given arrangement of dielectrics, D and E are parallel to the dielectric interface. Therefore, we can treat the capacitor as consisting of two capacitors C₁ and C₂ in parallel (same voltage across C₁ and C₂).
Thus,

and

The total capacitance/equivalent capacitance between terminals A and B is

QUESTION: 13

Three point charges -1 nC, 4 nC and 3 nC are located at (0, 0, 0) , (0,0,1) and (1,0,0,) respectively. To total energy contained in the system would be

A.
22.25 nJ
B.
15.56 nJ
C.
12.84 nJ
D.
13.37 nJ

Solution:

Total energy contained in the system is:
W=W₁+ W₂+ W₃
= 0 + Q₂V₂₁ + Q₃ (V₃₁ + V₃₂)

QUESTION: 14

The cylindrical coordinates equation

is called

A.
Laplace’s equation
B.
Poisson’s equation
C.
Larentz’s equation
D.
Maxwell’s equation

Solution:

QUESTION: 15

Image theory is applicable to problems involving

A.
electrostatic field only
B.
magnetostatic fields only
C.
both electrostatic and magnetostatic fields
D.
neither electrostatic nor magnetostatic fields

Solution:

QUESTION: 16

Match List-I with List-ll and select the correct answer using the codes given below the lists:
List-I
A. Joule’s law
B. Ohm’s law
C. Solenoidal current
D. Continuity equation

Codes:
A B C D
(a) 4 2 3 1
(b) 3 1 2 4
(c) 3 1 4 2
(d) 4 3 1 2

A.
(a)
B.
(b)
C.
(c)
D.
(d)

Solution:

QUESTION: 17

Assertion (A): An electric field produces no migration of charge in a dieiectric.
Reason (R): The permittivity of a dielectric is always less than the permittivity of vaccum.

A.
Both A and R are true and R is a correct explanation of A
B.
Both A and R are true but R is not a correct explanation of A
C.
A is true but R is false
D.
A is false but R is true

Solution:

Assertion is a true statement.
Reason is a false statement because the permittivity of dielectric is always more than the permittivity of vaccum which makes it a good insulator.
In a dielectric or insulator the electrons are so tightly bound by their parent nuclei in the equilibrium positions that they can not easily be detached by the applications of electric field. Due to this reason an electric field produces no migration of charge in a dielectric.

QUESTION: 18

The potential field at any point in space containing a dielectric material of relative permittivity 3 is given by V = (5x²y + 3yz² + 6xz) Volts, where x, y, z are in meters. The magnitude of volume charge density at point (1,2, 3) will be given by

A.
48 ∈₀
B.
96 ∈₀
C.
144 ∈₀
D.
192 ∈₀

Solution:

Given, V = (5x²y + 3yz² + 6xz) Volts and
∈_r = 3
Using Poisson’s equation,

Here,

= 10y + 0 + 6y = 16y
So, ρ_v = -∈ (10y+6y)
∴ ρ_v|_at (1,2,3) = -∈₀∈_r (10 x 2 + 6 x 2)
= -∈₀ x 3 x 32 = - 96∈₀
= 96∈₀ (Magnitude only)