Electric Charges and Fields Test

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Electric Charges and Fields Test - 25

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Question 1
1 / -0

Each of the two point charges are doubled and their distance is halved. Force of interaction becomes n times, where n is

A
1

B
4

C
16

D
$$\frac {1}{16}$$

Solution

By Coulomb's law, the force between two charges $$q_1$$ and $$q_2$$ is $$F=\dfrac{kq_1q_2}{r^2}$$ where $$r=$$ separation between charges.
When charges are doubled and their distance is halved , the force of interaction becomes $$F'=\dfrac{k(2q_1)(2q_2)}{(r/2)^2}=16 F$$
Thus, the value of $$n$$ will be $$16$$.
Question 2
1 / -0

What happens when the end of the paper clip of electroscope is touched by hands?

A
Electric current is produced

B
Foil strips remain as it is

C
Foil strips collapse

D
None of the above

Solution

When the paper clip of electroscope is touched by hands, the foil strips collapse because the charges on foil strips gets discharged through our body.
Question 3
1 / -0

When a charged body is brought near an electroscope:

A
the strips of foils opens up.

B
the strips of foils close.

C
the strips open and close simultaneously.

D
the strips neither open nor close.

Solution

When a charged body is brought near the top conducting disc of an electroscope, the disc gets induced with charges. These charges run down to the metal strips hanging from the paper clip. Now as both the strips get similar charge, those repel each other and thus open up.
Question 4
1 / -0

What will be the nature of charge on the metal paper clip of electroscope when a negatively charged body is brought in contact with it?

A
Positive

B
Negative

C
Partially positive

D
None of the above

Solution

The nature of charge will be negative. When a negatively charged body come in context with uncharged metal paper clip then some amount of negative change will transfer from changed body to metal clip. This phenomenon of change transfer is called Conduction.
Question 5
1 / -0

Electric switches and appliances should be tested only with

A
Right hand

B
Left hand

C
Both hands

D
Electric tester

Solution

We should not touch the electric appliance with bare hands. To check the supply in the circuit, the electric tester should be used.
Question 6
1 / -0

An electroscope is used:

A
to detect presence of charge on a body.

B
to determine the nature of charge on a body.

C
to measure the exact amount of charge on a body.

D
for both A and B.

Solution

Electroscope is a device which is use to detect whether a body is carrying charge and also to detect the nature of charge on the body.
Question 7
1 / -0

The electric field inside a hollow conducting sphere

A
Increases towards the centre

B
Decreases towards the centre

C
Is finite and constant throughout

D
Is zero

Solution

As there are no charges inside the hollow conducting sphere, as all charges reside on it surface. So, electric field inside the hollow conducting sphere is zero.
Question 8
1 / -0

The nature of charge on the object can be detected by:

A
Microscope

B
Electroscope

C
Electrometer

D
Telescope

Solution

A pre-charged electroscope detects the nature of charge on a charged body.
Question 9
1 / -0

The device, which can be used to detect whether an object is charged, is:

A
telescope

B
microscope

C
electroscope

D
none of the above

Solution

Electroscope is a device that is used for detection and identification of charge on objects.
Question 10
1 / -0

Two point charges $$+3\mu C$$ and $$+8\mu C$$ repel each other with a force of $$40 N$$. If a charge of $$-5\mu C$$ is added to each of them, the force between them will become:

A
$$-10 N$$

B
$$10 N$$

C
$$20 N$$

D
$$-20 N$$

Solution

Redistribution of charges takes place.
Charge $${ q }_{ 1 }=3\mu C$$
Charge $${ q }_{ 2 }=8\mu C$$
When third charge $${ q }_{ 3 }=-5\mu C$$ is added to each, then new charges on $${ q }_{ 1 }$$ and $${ q }_{ 2 }$$ will be
$${ q }_{ 1 }^{ \prime }=3-5=-2\mu C$$
and, $${ q }_{ 2 }^{ \prime }=8-5=3\mu C$$
Now,
In first case,
$$40=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cdot \dfrac { 3\times 8 }{ { r }^{ 2 } } $$
In second case,
$$F=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \times \dfrac { \left( -2\times 3 \right) }{ { r }^{ 2 } } $$
$$\therefore \dfrac { F }{ 40 } =\dfrac { -2\times 3 }{ 3\times 8 } $$
or $$F=-10N$$

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Electric Charges and Fields Test - 25

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Electric Charges and Fields Test - 25

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SHARING IS CARING

Question 1

1

4

16

$$\frac {1}{16}$$

Solution

Question 2

Electric current is produced

Foil strips remain as it is

Foil strips collapse

None of the above

Solution

Question 3

the strips of foils opens up.

the strips of foils close.

the strips open and close simultaneously.

the strips neither open nor close.

Solution

Question 4

Positive

Negative

Partially positive

None of the above

Solution

Question 5

Right hand

Left hand

Both hands

Electric tester

Solution

Question 6

to detect presence of charge on a body.

to determine the nature of charge on a body.

to measure the exact amount of charge on a body.

for both A and B.

Solution

Question 7

Increases towards the centre

Decreases towards the centre

Is finite and constant throughout

Is zero

Solution

Question 8

Microscope

Electroscope

Electrometer

Telescope

Solution

Question 9

telescope

microscope

electroscope

none of the above

Solution

Question 10

$$-10 N$$

$$10 N$$

$$20 N$$

$$-20 N$$

Solution

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