Electric Charges and Fields Test

Result

Electric Charges and Fields Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Which is the best conductor of electricity?

A
Iron

B
Silver

C
Copper

D
Carbon

Solution

Silver is a best conductor of electricity because it contains maximum number of free electron.Also the resistance of silver is very low.

Option B is correct.
Question 2
1 / -0

Two positive point charges $$q_1$$ and $$q_2$$ are kept at the distance d. If the distance between them is triple. Calculate by which factor new electrostatic force is decreased.

A
3

B
6

C
8

D
9

E
12

Solution

By coulomb's law, the force between the charges is $$F=\dfrac{kq_1q_2}{r^2}$$
When $$r=3r, F'=\dfrac{kq_1q_2}{9r^2}=F/9$$
Question 3
1 / -0

What would happen to the electrostatic force between a pair of charged particles if both charges were tripled and the distance between them were also tripled?

A
It would decrease by a factor of $$4$$

B
It would decrease by a factor of $$2$$

C
It would remain unchanged

D
It would increase by a factor of $$2$$

E
It would increase by a factor of $$4$$

Solution

The electric force between two charges $$q_1$$ and $$q_2$$ is $$F=\dfrac{kq_1q_2}{r^2}$$ where r be the distance between them.
Now, $$F'=\dfrac{kq'_1q'_2}{r'^2}=\dfrac{k(3q_1)(3q_2)}{(3r)^2}=\dfrac{kq_1q_2}{r^2}=F$$
Question 4
1 / -0

Two point charges with charges $$3$$ micro coulombs and $$4$$ micro coulombs are separated by $$2$$ cm.The value of the force between them?

A
$$600N$$

B
$$300N$$

C
$$540N$$

D
$$270N$$

E
$$400N$$

Solution

Given : $$Q_1 = 3 \times 10^{-6} C$$ $$Q_2 = 4\times 10^{-6} C$$ $$r =2 cm = 0.02$$ $$m$$
Coulomb force between them $$F = \dfrac{K Q_1Q_2}{r^2}$$ where $$K = \dfrac{1}{4\pi \epsilon_o} = 9 \times 10^9\ Nm^2C^{-2}$$
$$\therefore$$ $$F = \dfrac{9 \times 10^9 \times (3 \times 10^{-6}) \times (4 \times 10^{-6})}{(0.02)^2} =270$$ $$N$$
Question 5
1 / -0

Which of the following materials has a negative temperature coefficient of resistance?

A
Copper

B
Aluminium

C
Carbon

D
brass

Solution

Carbon has a negative temperature coefficient of resistance.
Option C is correct.
Question 6
1 / -0

When we wear nylon dresses during winter then there is ______ current which gets produced due to contact with out body. Fill in the Blank.

A
Magnetic

B
Electrostatic

C
Potential

D
kinetic

Solution

When we wear nylon dresses during winter then there is electrostatic current which gets produced due to contact with out body.

Option B is correct.
Question 7
1 / -0

A closed switch has a resistance of :

A
zero ohm

B
about $$50$$ ohms

C
about $$500$$ ohms

D
infinity

Solution

A closed switch has a Zero resistance.

Option A is correct.
Question 8
1 / -0

When we rub our comb with hair and then take near to the paper pieces, all pieces get stuck to the comb. This happens due tyo which phenomena?

A
Electrostatic Induction

B
Magnetic effect

C
Potential effect

D
Kinetic effect

Solution

When negative charges is created in the comb by rubbing it with hair and pieces of paper are exposed to the electric field created by the charge, the part of paper closer to the comb will be positively charged due to electromagnetic induction, and the farther would be negatively charged, causing the paper to be attracted to the comb.
Question 9
1 / -0

Identify which of the following diagrams best represents the electric field lines around a positive charge?

A

B

C

D

E

Solution

Electric field lines originate at the positive charge and diverge (spread) in all directions into the space and end at the negative charge.
Thus option B is correct.
Question 10
1 / -0

A test charge q is located 1 meter from a much larger and stationary charge Q. While at this location, the test charge q, experiences a force of F from the stationary charge Q. The test charge is then moved to a new location 2 meters from Q. What force will the test charge q experience from the stationary charge Q at the new location?

A
(1/4) F

B
1/2 F

C
F

D
2F

Solution

Given : $$r_i = 1$$ m $$r_f = 2$$ m
Initial electric force on the test charge $$F = \dfrac{kQq}{r_i^2} = \dfrac{kQq}{1^2} = kQq$$

Final electric force on the test charge $$F' = \dfrac{kQq}{r_f^2} = \dfrac{kQq}{2^2} = \dfrac{F}{4}$$