Electric Charge, Coulomb's law and Electric Field Test

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Electric Charge, Coulomb's law and Electric Field Test - 10

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Weekly Quiz Competition

Question 1
1 / -0.25

One quantum of charge should be at least be equal to the charge in coulomb:

A
1.6 ×10^-¹⁷

B
1.6 ×10^-1⁹

C
1.6 ×10^-¹⁰

D
4.8 ×10^-¹⁰

Solution

Quantum of charge is represented by q=ne... where e is the charge of an electron. for 1 quantum of charge the value is 1.6 ×10^-19 C
Question 2
1 / -0.25

The force between two charges is 120 N. If the distance between the charges is doubled, the force will be

A
60 N

B
30 N

C
40 N

D
15 N

Solution
Question 3
1 / -0.25

Which one of the following statement regarding electrostatics is wrong ?

A
Charge is quantized

B
Charge is conserved

C
There is an electric field near an isolated charge at rest

D
A stationary charge produces both electric and magnetic fields

Solution

A stationary charge doesn 't produce a magnetic field , it only produces an electric field.
Question 4
1 / -0.25

Two similar charge of +Q , as shown in figure are placed at A and B. _q charge is placed at point C midway between A and B. _q charge will oscillate if

A
It is moved towards A

B
It is moved towards B

C
It is moved upwards AB

D
Distance between A and B is reduced

Solution

When -q is displaced left or right of the force is increased so -q move towards that force hence will not perform SHM but when its displaced to AB it will perform SHM because it will return to its mean position.
Question 5
1 / -0.25

When the distance between two charged particle is halved, the force between them becomes

A
One fourth

B
One half

C
Double

D
Four times

Solution

Earlier force between two particles is
F=kq¹ q² / r²
When distance is halved force becomes
F ′= kq₁ q² /(r/2)² =4kq₁ q₂ /r² =4F
Question 6
1 / -0.25

Two point charges in air at a distance of 20 cm. from each other interact with a certain force. At what distance from each other should these charges be placed in oil of relative permittivity 5 to obtain the same force of interaction

A
8.94 ×10^-2 m

B
0.894 ×10^-² m

C
89.4 ×10^-² m

D
8.94 ×10² m

Solution

d=8.94 cm or
d=8.94x10^-2
Question 7
1 / -0.25

A certain charge Q is divided at first into two parts, (q) and (Q-q). Later on the charges are placed at a certain distance. If the force of interaction between the two charges is maximum then

A
(Q/q) = (4/1)

B
(Q/q) = (2/1)

C
(Q/q) = (3/1)

D
(Q/q) = (5/1)

Solution

The force between q &Q-q is given by :-
F = kq(Q-q)/r ²
As it is given that force between these two is Maximum....so it will be max. when dF/dq is zero
i.e, dF/dq = {k(Q-2q)}/r ²= 0
so, Q = 2q
Q/q = 2/1
Question 8
1 / -0.25

Two small balls having equal positive charge Q (Coulomb) on each are suspended by two insulating strings of equal length 'L 'metre, from a hook fixed to a stand. The whole set up is taken in a satellite in to space where there is no gravity (state of weight lessness). Then the angle (q) between the two strings is

A
0 º

B
90 º

C
180 º

D
0º < q < 180º

Solution

There is a condition of weightlessness in a satellite.
Therefore, mg=0
Due to electrostatic force of repulsion between the balls, the string would become horizontal.
Therefore, angle between the string =180
Question 9
1 / -0.25

Two point charges -1 ×10^-6 C and 4 ×10^-6 C separated by 3 m distance have zero field location of

A
2 m

B
4 m

C
3 m

D
1 m

Solution

x=l/[ √(q₂ /q₁ )-1]
=3/√[(4x10^-6 /1x10^-6 )-1]
= (3/2)-1
=3
So, 3m away from -1x10^-6 C
Question 10
1 / -0.25

Two charges 4q and q are placed 30 cm. apart. At what point the value of electric field will be zero

A
10 cm. away from q and outside the line joining them.

B
10 cm. away from 4q and out side the line joining them.

C
20 cm. away from 4q and between the charge

D
None

Solution

Where E=0 this point is called neutral point.
it is the point where electric field of both charge is same
we know tha t E=kQ/r^2
here k=1/4pi €
for 4q charge
let at ''a ''distance we get E=0 which is from q charge
so distance from 4q of 'a 'point is 30-a
electric field by 4q charge on a is
E=k4q/(30-a)^2
electric field by q charge on a point
E=kq/a^2
both electric field are equal so put them equal
k4q/(30-a)^2=kq/a^2
solve this we get
4a^2=(30-a)^2
2a=30-a
3a=30
a=10
so at a distance 10cm from charge q we get E=0
distance from 4q charge 30-10=20cm.

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Electric Charge, Coulomb's law and Electric Field Test - 10

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Electric Charge, Coulomb's law and Electric Field Test - 10

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Question 1

1.6 ×10-17

1.6 ×10-19

1.6 ×10-10

4.8 ×10-10

Solution

Question 2

60 N

30 N

40 N

15 N

Solution

Question 3

Charge is quantized

Charge is conserved

There is an electric field near an isolated charge at rest

A stationary charge produces both electric and magnetic fields

Solution

Question 4

It is moved towards A

It is moved towards B

It is moved upwards AB

Distance between A and B is reduced

Solution

Question 5

One fourth

One half

Double

Four times

Solution

Question 6

8.94 ×10-2 m

0.894 ×10-2 m

89.4 ×10-2 m

8.94 ×102 m

Solution

Question 7

(Q/q) = (4/1)

(Q/q) = (2/1)

(Q/q) = (3/1)

(Q/q) = (5/1)

Solution

Question 8

0 º

90 º

180 º

0º < q < 180º

Solution

Question 9

2 m

4 m

3 m

1 m

Solution

Question 10

10 cm. away from q and outside the line joining them.

10 cm. away from 4q and out side the line joining them.

20 cm. away from 4q and between the charge

None

Solution

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1.6 ×10^-¹⁷

1.6 ×10^-1⁹

1.6 ×10^-¹⁰

4.8 ×10^-¹⁰

8.94 ×10^-2 m

0.894 ×10^-² m

89.4 ×10^-² m

8.94 ×10² m