Electric Charges and Fields Test

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Electric Charges and Fields Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

On doubling the charge on one particle while leaving all other factors the same will:

A
cause the force to increase by a factor of 2

B
cause the force to decrease by a factor of 4

C
cause the force to increase by a factor of 4

D
cause the force to remain the same

E
cause the force to decrease by a factor of 2

Solution

Coulomb force between two charges $$F = \dfrac{kq_1q_2}{r^2}$$
Now the charge of one particle is doubled i.e $$q_1' = 2q_1$$
$$\therefore$$ New coulomb force between them $$F' = \dfrac{k (2q_1)q_2}{r^2} = 2F$$
Thus the force increases by a factor of 2.
Question 2
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Two charges $$q_1$$ and $$q_2$$, are separated by a distance r and apply a force F to each other. If both charges are doubled and the distance between them is also doubled, find out the new force between them ?

A
$$\dfrac{1}{2}F$$

B
$$\dfrac{1}{3}F$$

C
$$F$$

D
$$8F$$

E
$$16F$$

Solution

Initial force between the charges $$F = \dfrac{kq_1q_2}{r^2}$$
Now distance between charges is doubled and both the charges are also doubled i.e $$q'_1 = 2q_1$$, $$q_2' = 2q_2$$ and $$r' = 2r$$
Thus new force $$F' = \dfrac{k (2q_1) (2q_2)}{(2r)^2} = F$$
Question 3
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Calculate the magnitude of the force between two electrons which are separated by a distance of $$10^{-10}m$$. If the magnitude of the charge is $$10^{-19}C$$ and Coulomb constant is $$10^{10}N\;m^{2}/C^{2}$$.

A
$$10^{-8}N$$

B
$$10^{-28}N$$

C
$$10^{-38}N$$

D
$$10^{-48}N$$

E
$$10^{-58}N$$

Solution

Given : $$Q = 10^{-19} C$$ $$k = 10^{10}$$ $$r = 10^{-10}$$ m
Force between them $$F = \dfrac{kQ^2}{r^2} = \dfrac{10^{10} \times 10^{-19} \times 10^{-19}}{(10^{-10})^2} = 10^{-8}$$ N
Question 4
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If the electric force between two unknown charges is attractive, identify which of the following statement must be true?

A
One charge is positive and the other charge is negative

B
Both charges are positive

C
Both charges are negative

D
The two charges must be equal in magnitude

E
The force must be directed towards the larger charge

Solution

The direction of electric force between the charges depends upon the nature of the charges.
When the charges are of same nature, that is, both positive or both negative, the force between them is repulsive in nature.
When the charges are of opposite nature, that is, one positive and one negative, the force between them is attractive in nature.
Question 5
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Find out on which of following factors does strength of the force between charged particles depend.

A
the sign of the charges

B
the mass of the charges

C
the orientation of the charges

D
the direction of the force

E
the distance between charges

Solution

Coulomb force between charged particles $$F = \dfrac{k q_1q_1}{r^2}$$
Thus the strength of force depends on the distance between the charged particles.
Question 6
1 / -0

On doubling the distance between particles while leaving all other factors the same will:

A
cause the force to decrease by a factor of $$\dfrac{1}{4}$$

B
cause the force to decrease by a factor of $$2$$

C
cause the force to increase by a factor of $$4$$

D
cause the force to increase by a factor of $$2$$

E
cause the force to remain the same

Solution

Let the initial distance between the charges be $$r$$.
Thus initial coulomb force $$F = \dfrac{kq_1q_2}{r^2}$$
Now the distance between them is doubled i.e $$r' = 2r$$
Thus new coulomb force between them $$F = \dfrac{kq_1q_2}{(2r)^2} =\dfrac{F}{4}$$
Hence the force decreases by a factor of $$1/4$$.
Question 7
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A positive charged particle is kept on the plane surface. There are two points X and Y lies on the plane surface as shown above. If the point Y is three times as far away from Q as point X. What is the ratio of the electric force that would act on a small charge placed at point Y compared to the charge placed at point X?

A
$$1 :9$$

B
$$1 : 3$$

C
$$1 : 1$$

D
$$3 : 1$$

E
$$9 : 1$$

Solution

Given : $QX = r$$ $$QY = 3r$$
Let the positive charge at point Q be $$q$$.
Electric field due to $$q$$ at point X $$E_x = \dfrac{kq}{r^2}$$

Electric field due to $$q$$ at point Y $$E_y = \dfrac{kq}{(3r)^2} = \dfrac{kq}{9r^2}$$
$$\implies$$ $$\dfrac{E_y}{E_x} = \dfrac{1}{9}$$
Question 8
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Electric lines of force about a negative point charge are:

A
Circular anti-clockwise

B
Circular clockwise

C
Radially inwards

D
Radially outwards

Solution

Hint:- Use the properties of electric field lines.
Explanation:-
$$\bullet$$ Electric field lines are always originated from positive charge and end at negative charge.
$$\bullet$$ Electric field lines are perpendicular to the surface of point charge.
So for a negative point charge electric field lines are radially inwards.

Hence option C is correct
Question 9
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A charge "Q" and "2Q" are 0.05 meters apart and isolated. The ratio of the electrostatic force on the charge Q to the force on charge 2Q is:

A
2 : 1

B
1 : 1

C
1 : 2

D
1 : 4

Solution

According to Newton's 3rd law, every action has equal and opposite reaction.
Thus the force exerted by charge on $$Q$$ due to $$2Q$$ is equal but opposite in direction to the force exerted by charge on $$2Q$$ due to $$Q$$ i.e $$F_Q = F_{2Q}$$
$$\implies$$ $$\dfrac{F_Q}{F_{2Q}} = 1$$
Question 10
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Electroscope is used

A
to detect and test small electric charges

B
to calculate the amount of electric charge flowing through the conductor in the given interval of time

C
to find out the presence of antimatter

D
to test the presence of magnetic field

Solution

Electroscope is an old electrical instrument , used to detect the presence and magnitude of electric charges on a body .

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Electric Charges and Fields Test - 28

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Question 1

cause the force to increase by a factor of 2

cause the force to decrease by a factor of 4

cause the force to increase by a factor of 4

cause the force to remain the same

cause the force to decrease by a factor of 2

Solution

Question 2

$$\dfrac{1}{2}F$$

$$\dfrac{1}{3}F$$

$$F$$

$$8F$$

$$16F$$

Solution

Question 3

$$10^{-8}N$$

$$10^{-28}N$$

$$10^{-38}N$$

$$10^{-48}N$$

$$10^{-58}N$$

Solution

Question 4

One charge is positive and the other charge is negative

Both charges are positive

Both charges are negative

The two charges must be equal in magnitude

The force must be directed towards the larger charge

Solution

Question 5

the sign of the charges

the mass of the charges

the orientation of the charges

the direction of the force

the distance between charges

Solution

Question 6

cause the force to decrease by a factor of $$\dfrac{1}{4}$$

cause the force to decrease by a factor of $$2$$

cause the force to increase by a factor of $$4$$

cause the force to increase by a factor of $$2$$

cause the force to remain the same

Solution

Question 7

$$1 :9$$

$$1 : 3$$

$$1 : 1$$

$$3 : 1$$

$$9 : 1$$

Solution

Question 8

Circular anti-clockwise

Circular clockwise

Radially inwards

Radially outwards

Solution

Question 9

2 : 1

1 : 1

1 : 2

1 : 4

Solution

Question 10

to detect and test small electric charges

to calculate the amount of electric charge flowing through the conductor in the given interval of time

to find out the presence of antimatter

to test the presence of magnetic field

Solution

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