Electric Charges and Fields Test

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Electric Charges and Fields Test - 5

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Question 1
4 / -1

The force per unit charge is known as

A
electric flux

B
electric field

C
electric potential

D
electric current

Solution

The force per unit charge is known as: E= q/F
This formula f is hold for electric field.
Question 2
4 / -1

Electrical as well as gravitational affects acan be thought to be caused by fields. Which the following is true of an electrical or gravitational field?

A
The field concept is often used to describe contact forces.

B
Gravitational or electric field does not exit in the space around an object.

C
Fields are useful for understanding forces acting through a distance.

D
There is no way to verify the existence of a force field since it is just a concept.

Solution

In physics, concept of field is a model used to explain the influence that a massive body or charged particle extends into the space around itself, producing a force on another massive body or charged body placed in space. Thus, concept of field is used to explain gravitational and electrostatic phenomena.
Question 3
4 / -1

The electric field at a point is

A
always continuous

B
continuous if there is no charge at that point

C
discontinuous if there is a charge at that point

D
both (b) and (c) are correct

Solution

Electric field at a point is continuous if there is no charge at that point. And the field is discontinuous if there is charge at that point. So both options (b) and (c) are correct.
Question 4
4 / -1

The dimensional formula of electric intensity is

A
[M¹L¹T³A^-1]

B
[ML^-1T^-3A¹]

C
[M¹L¹T^-3A^-1]

D
[M¹L²T¹A¹]

Solution
Question 5
4 / -1

If the charge on an object is doubled then electric field becomes

A
half

B
double

C
unchanged

D
thrice

Solution

As
If q ' = 2q, then E' = E' = 2E
So electric field is doubled.
Question 6
4 / -1

A force of 2.25 N acts on a charge of 15 x 10^-4 C. The intensity of electric field at that point is

A
150 N C^-1

B
15 N C^-1

C
1500 N C^-1

D
1.5 N C^-1

Solution
Question 7
4 / -1

A conducting sphere of radius 10 cm has unknown charge. If the electric field at a distance 20 cm from the centre of the sphere is 1.2 x 10³N C^-1 and points radially inwards. The net charge on the sphere is

A
-4.5 x 10^-9 C

B
4.5 x 10⁹ C

C
-5.3 x 10^-9 C

D
5.3 x 10⁹ C

Solution

Here, distance of point from the centre of the sphere, r = 20 cm = 0.2m
Electric field, E = -1.2 x 10³ N C^-1
Question 8
4 / -1

A particle of mass 10^-3kg and charge 5μC is thrown at a speed of 20m s^-1 against a uniform electric field of strength 2 x 10⁵ N C^-1. The distance travelled by particle before coming to rest is

A
0.1 m

B
0.2 m

C
0.3 m

D
0.4 m

Solution

F = qE = 5 x 10^-6 x 2 x 10⁵ = 1 N
Since, the particle is thrown against the field
∴ a = -F/m = -(1/10^-3) = 10³ms^-2
As v² - u² = 2as
∴ 0² - (20)² = 2 x (-10³) x s or s = 0.2 m
Question 9
4 / -1

An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of magnitude 2 x 10⁴N/C. The time taken by the electron to fall this distance is

A
1.3 x 10² s

B
2.1 x 10^-12 s

C
1.6 x 10^-10 s

D
2.9 x 10^-9 s

Solution

In figure the field is upward. So the negatively charged electron experiences a downward force.

∴ The acceleration of electron is
a_c = eE/m_e
The time required by the electron to fall through a distance h is
Question 10
4 / -1

The electric field that can balance a charged particle of mass 3.2 x 10^-27 kg is (Given that the charge on the particle is 1.6 x 10^-19 C)

A
19.6 x 10^-8 N C^-1

B
20 x 10^-6 N C^-1

C
19.6 x 10⁸ N C^-1

D
20 x 10⁶ N C^-1

Solution

Here, m = 3.2 x 10^-27kg, e = 1.6 x 10^-19 C, g = 9.8 m s^-2
Question 11
4 / -1

An oil drop of 10 excess electrons is held stationary under a constant electric field of 3.65 x 10⁴ N C^-1 in Millikan's oil drop experiment. The density of oil is 1.26 g cm^-3. Radius of the oil drop is
(Take, g = 9.8 m s^-2, e = 1.6 x 10^-19 C)

A
1.04 x 10^-6 m

B
4.8 x 10^-5 m

C
4.8 x 10^-18 m

D
1.13 x 10^-18 m

Solution

Here, n = 10, E = 3.65 x 10⁴N C^-1
ρ_oil = 1.26gcm^-3 = 1.26 x 10³ kg m^-3
As the droplet is stationary weight of droplet = force due to electric field
or (4/3)πr³ρg = neE ∴ r³ = 3neE/4πρg

or r = (1.13 x 10^-18)^1/3= 1.04 x 10^-6 m
Question 12
4 / -1

Five equal charges each of value q are placed at the corners of a regular pentagon of side a. The electric field at the centre of the pentagon is

A

B

C

D
zero

Solution

The electric field at the centre of pentagon would be zero.
Question 13
4 / -1

Five equal charges each of value q are placed at the corners of a regular pentagon of side a. What will be the electric field at centre O, if the charge from one of the corners (say A) is removed?

A
(q/4πε₀r²) along OA

B
(2q/4πε₀r²) along OB

C
(q²/4πε₀r²) along OC

D
(2q/4πε₀r²) along OA

Solution

When a charge q from corner A is removed, electric field at O is E₁ = (q/4πε₀r²) along OA
Question 14
4 / -1

In question number 45, what will be the electric field at O if the charge q at A is replaced by -q?

A
(q/4πε₀r²) along OB

B
(2q/4πε₀r²) along OA

C
(4q²/4πε₀r²) along OC

D
zero

Solution

If the charge q at A is replaced by -q, it is equivalent to adding charge -2q at A. Therefore, electric field at O, E₂ = (2q/4πε₀r²) along OA.
Question 15
4 / -1

The tracks of three charged particles in a uniform electrostatic field as shown in the figure. Which particle has the highest charge to mass ratio?

A
A

B
B

C
C

D
A and B

Solution

Particles A and B have negative charges because they are being deflected towards the positive plate of the electrostatic field. Particle C has positive charge because it is being deflected towards the negative plate.
∴ Deflection of charged particle in time t in y-direction
h = 0 x t + (1/2)at² = (1/2)(qE/m)t² i.e., h ∝ q/m
As the particle C suffers maximum deflection in y-direction, so it has highest charge to mass q/m ratio.