Electric Charge, Coulomb's law and Electric Field Test

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Electric Charge, Coulomb's law and Electric Field Test - 4

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Weekly Quiz Competition

Question 1
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When a negatively charged conductor is connected to earth,

A
No charge flow occurs.

B
Protons flow from the conductor to the earth.

C
Electrons flow from the earth to the conductor.

D
Electrons flow from the conductor to the earth.
Solution
Explanation:
When a negatively charged conductor is connected to the earth, electrons will flow from the conductor to the earth. This is because electrons have a negative charge and they will be repelled from the negatively charged conductor and attracted to the positively charged earth. As electrons flow from the conductor to the earth, the negative charge on the conductor will gradually decrease until it becomes neutral.
Option A is incorrect because charge flow does occur when a negatively charged conductor is connected to the earth.
Option B is incorrect because protons have a positive charge and they are not free to move in a conductor.
Option C is incorrect because electrons flow from the earth to the conductor, not the other way around.
Question 2
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A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

A
1.7 ×10⁵ Nm² /C

B
2.1 ×10⁵ Nm² /C

C
2.2 ×10⁵ Nm² /C

D
1.9 ×10⁵ Nm² /C

Solution
Question 3
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Two identical conductors of copper and aluminium are placed in an identical electric field. The magnitude of induced charge in the aluminium will be:

A
Zero

B
Equal to that of copper

C
Greater than in copper

D
less than in copper

Solution

As aluminium and copper are metals, their mobile electrons move under the influence of the external field until they reach the surface of the metal and collect there. External electric fields induce surface changes on metal objects that exactly cancel the field within. Since the field applied is same in both case, the induced charge will be the same.
Hence, the magnitude of induced charge in the aluminium will be the equal to that of copper .
Question 4
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A point charge causes an electric flux of −1.0 ×10³ Nm² /C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?

A
10³ Nm² /C,−7.8nC

B
10³ Nm² /C,−8.8nC

C
−10³ Nm² /C,−8.8nC

D
−10³ Nm² /C,−6.8nC

Solution

Electric flux, Φ = −1.0 ×10³ N m² /C
Radius of the Gaussian surface,
r = 10.0 cm
Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., −10³ N m² /C.
Electric flux is given by the relation,
q ϕ=q ∈0
Where,
q = Net charge enclosed by the spherical surface
∈₀ = Permittivity of free space = 8.854 ×10⁻¹² N⁻¹ C² m⁻²
∴ qq=ϕ∈0
= −1.0 ×10³ ×8.854 ×10⁻¹²
= −8.854 ×10⁻⁹ C
= −8.854 nC
Therefore, the value of the point charge is −8.854 nC.
Question 5
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A hollow spherical conductor of radius 2m carries a charge of 500 μ C. Then electric field strength at its surface is

A
4.5 ×10⁶ N/C

B
Zero

C
2.25 ×10⁶ N/C

D
1.125 ×10⁶ N/C .

Solution
Question 6
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Two point charges A and B, having charges +Q and –Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes

A
4F/3

B
F

C
9F/16

D
16F/9

Solution

In case I :

In Case II :

From equations (i) and (ii),
Question 7
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A

B

C

D

Solution

Let m be mass of each ball and q be charge on each ball. Force of repulsion,

In equilibrium
Tcosq = mg ...(i)
Tsinq = F ...(ii)
Divide (ii) by (i), we get,

From figure (a),

Divide (iv) by (iii), we get
Question 8
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Under the influence of the coulomb field of charge +Q, a charge −q is moving around it in an elliptical orbit. Find out the correct statement(s).

A
The linear momentum of the charge −q is constant

B
The angular velocity of the charge −q is constant

C
The angular momentum of the charge −q is constant.

D
The linear speed of the charge −q is constant

Solution

Since the charge –q is moving in elliptical orbit so to make its motion stable the total angular momentum of the charge is constant since it experience a centripetal force from the charge +Q so it follow the motion as the motion of earth around sun.
Question 9
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A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field . Due to the force , its velocity increases from 0 to 6 m s –1 in one second duration. At that instant the direction of the field is
reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively

A
2 m s^–1 , 4 m s^–1

B
1 m s^–1 , 3 m s^–1

C
1 m s^–1 , 3.5 m s^–1

D
1.5 m s^–1 , 3 m s^–1

Solution

Acceleration
Question 10
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Three point charges +q, –2q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are

A
√2qa along the line joining points (x = 0, y = 0,
z = 0) and (x = a, y = a, z = 0)

B
qa along the line joining points (x = 0, y = 0,
z = 0) and (x = a, y = a, z = 0)

C
√2qa along + x direction

D
√2qa along + y direction

Solution

This consists of two dipoles, –q and +q with dipole moment along with the +y-direction and –q and +q along the x-direction.

Along the direction 45 °that is along OP, where P is (+a, +a, 0).