Electric Charges and Fields Test

Result

Electric Charges and Fields Test - 32

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Two small objects X and Y are permanently separated by a distance 1 cm. Object X has a charge of $$+1.0\mu C$$ and object Y has a charge of $$-1.0\mu C$$. A certain number of electrons are removed from X and put onto Y to make the electrostatic force between the two objects an attractive force whose magnitude is 360 N. Number of electrons removed is :

A
$$8.4\times 10^{13}$$

B
$$6.25\times 10^{12}$$

C
$$4.2\times 10^{11}$$

D
$$3.5\times 10^{10}$$

Solution

New attractive force is -360 N.

$$- 360 \ N=\dfrac {K(q_1)(q_2)}{r^2}$$

$$K=9\times 10^9$$, $$r=10^{-2}\ m$$

$$- 360 \ N=\dfrac {K(1\times 10^{-6}+ne)(-(1\times 10^{-6})-ne)}{(1\times 10^{-2})^2}$$

$$\dfrac {360\times 10^{-4}}{9\times 10^9}=(1\times 10^{-6}+ne)^2$$

$$\dfrac {4}{10^{12}}=(1\times 10^{-6}+ne)^2$$

$$ {2\times 10^{-6}}=1\times 10^{-6}+ n e.$$

$$n e =1\times 10^{-6}$$

$$n= {1\times 10^{-6}}=\dfrac {1\times 10^{-6}}{1\cdot 6\times 10^{-19}}=6\cdot 25\times 10^{12}$$
Question 2
1 / -0

Two particles with identical positive charges and a separation of $$2.5\times 10^{-2}m$$ are released from rest. Immediately after their release, particle 1 has acceleration whose magnitude is $$4\times 10^{3}ms^{-2}$$. Particle 1 has a mass of $$9\times 10^{-6}kg$$. Then each of the particle has charge of :

A
$$3\times 10^{-8}C$$

B
$$4\times 10^{-8}C$$

C
$$5\times 10^{-8}C$$

D
$$6\times 10^{-8}C$$

Solution

Force on particle 1 = mass X acceleration =$$ 9\times 10^{-6}kg \times4\times10^{3}\ ms^{-2} = 0.036\ N$$

By Coulomb's law, this force $$= \dfrac{1}{4\pi\varepsilon o}\dfrac{q^{2}}{r^{2}}$$, where $$q$$ is the charge on the particles and $$r$$ is the distance between them $$(=2.5\times 10^{-2}m).$$

$$\therefore \dfrac{1}{4\pi\varepsilon o}\dfrac{q^{2}}{r^{2}} = 0.036\ N$$

$$\Rightarrow 9\times 10^{9}\times \dfrac{q^{2}}{2.5\times 10^{-2}\times2.5\times10^{-2}} = 0.036\ N$$

$$\Rightarrow q = 5\times 10^{-8}\ C$$
Question 3
1 / -0

A point charge $$Q_{1}$$ exerts some force on a second point charge $$Q_{2}$$. If a 3rd point charge $$Q_{3}$$ is
brought near, the force of $$Q_{1}$$ exerted on $$Q_{2}$$(Without changing their respective positions):

A
Will increase

B
Will decrease

C
Will remain unchanged

D
Will increase if $$Q_{3}$$ is of the same sign as $$Q_{1}$$ and will decrease if $$Q_{3}$$ is of opposite sign

Solution

As the principal of superposition states that the force exerted by a charge on another point is independent of the presence of the other charges and the net force on a point charge due to the number of charges will be the vector sum of the forces exerted by the other charges.

Therefore, despite of the presence of the third charge $$Q_3$$, the force exerted by the charge $$Q_2$$ by the charge $$Q_1$$ will remain unchanged.
However, the net force experienced by the charge $$Q_2$$ will change due to the presence of $$Q_3$$.
Question 4
1 / -0

If two point charges of 1 coulomb each are placed 1 km apart in vaccum, the force between them will be:

A
$$9\times 10^3N$$

B
$$9\times 10^{-3}N$$

C
$$1.1\times 10^{-4}N$$

D
$$10^{-6}N$$

Solution

$$q_1=q_2=1C$$
$$d=1 k =10^3 m$$
$$F=\dfrac {Kq_1q_2}{d^2}=\dfrac {9\times 10^9\times 1\times 1}{(10^3)^2}$$
$$=9\times 10^9\times 10^{-6}$$
$$=9\times 10^3N$$
Question 5
1 / -0

Electric lines of force are:

A
Exist everywhere

B
Exist only in the immediate vicinity of electric charges

C
Exist only when both positive and negative charges are near one another

D
Are imaginary

Solution

Electric lines of forces are imaginary lines it shows that if a point charge is put up somewhere than its direction of motion can be shown by it.
Question 6
1 / -0

A charge of $$-30 \mu C$$ is placed at 50 cm from another charge of $$+25\mu C$$ is vacuum. The force between them is:

A
$$60 N$$

B
$$27 N$$

C
$$90 N$$

D
$$120 N$$

Solution

$$q_1=-30\mu C$$
$$q_2=25\mu C$$
$$d=50 cm$$ or $$0.5 m$$

$$\displaystyle \therefore F=\dfrac {9\times 10^9\times (30)\times (25)\times 10^{-12}}{(0.5)^2}$$

$$\displaystyle =\dfrac {9\times 750\times 10^{-3}}{0.25}=\dfrac {9\times 75}{25}=27N$$
this force will be attractive in nature
Question 7
1 / -0

If the force between two charged objects is to be left unchanged, even though the charge on one of the objects is halved, keeping the other the same, the original distance of separation 'd' should be changed to:

A
$$d/\sqrt 2$$

B
$$d/2$$

C
$$\sqrt 2d$$

D
2 d

Solution

$$F\alpha \dfrac {q_1q_2}{d^2}$$
Given $$q_1=q_1/2; q_2=q_2$$
$$\therefore F'=k\dfrac {q_1q_2}{2d'^2}$$
$$\displaystyle \dfrac {kq_1q_1}{2(d^1)^2}=\dfrac {kq_1q_2}{d^2}$$
$$d^2=2(d^1)^2$$
$$\dfrac {d}{\sqrt 2}=d'$$
Question 8
1 / -0

Two particles having charges $$q_1$$ and $$q_2$$ when kept at a certain distance, exert a force F on each other. If the distance between the two charges is reduced to half then the force exerted on each other would be:

A
2 F

B
4 F

C
8 F

D
16 F

Solution

Given $$F=\dfrac {Kq_1q_2}{d^2}$$
If $$d^1=\dfrac {d}{2}$$

$$F^1=\dfrac {Kq_1q_2}{\dfrac {d^2}{4}}$$

$$F^1=4 F$$
Question 9
1 / -0

If the force between two charged objects separated by a distance d is to be left unchanged even though the charge on one of the objects is halved, keeping other the same, the new distance of separation becomes :

A
$$d\sqrt { 2 } $$

B
$$\dfrac { d }{ \sqrt { 2 } } $$

C
$$\dfrac { \sqrt { 2 } }{ d } $$

D
$$2d$$

Solution

According to coulomb's inverse square law,

$$ F = \dfrac{kq_{1}q_{2}}{d^2}$$

If charge on one of the objects is halved, but force remains unchanged,

$$F= \dfrac{\dfrac{kq_{1}}{2} q_{2}}{{d_1}^2}$$

Hence, $$\dfrac{q_1}{d^2}= \dfrac{\dfrac{q_1}{2}}{{d_1}^2}$$

$$\therefore d_1 = \dfrac{d}{\sqrt 2 }$$
Question 10
1 / -0

If two electrons are separated by a distance of $$0.5\mathring { A }$$, then the ratio of coulomb force and weight of the electron is approximately :

A
$${ 10 }^{ 16 }$$

B
$${ 10 }^{ 18 }$$

C
$${ 10 }^{ 20 }$$

D
$${ 10 }^{ 22 }$$

Solution

Coulomb force between the electrons is given by, $$[F = (k \times e^{2})/r^{2}]$$
$$[F = [9 \times 10^{9} \times (1.6 \times 10^{-19})^{2}]/(0.5 \times 10^{-10})^{2}]$$ N.
$$[F = 9.2 \times 10^{-8}]$$ N
weight of an electron is given by, $$[W = m \times g]$$
$$[W = 9.1\times 10^{-31} \times 9.8]$$ N
$$[W = 8.9 \times 10^{-30}]$$ N
Now , $$[F/W = 9.2\times 10^{-8}/8.9 \times 10^{-30}]$$
$$[F/W = 1.03\times 10^{22}]$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Electric Charges and Fields Test - 32

Result

Electric Charges and Fields Test - 32

Score

-

Rank

-

SHARING IS CARING

Question 1

$$8.4\times 10^{13}$$

$$6.25\times 10^{12}$$

$$4.2\times 10^{11}$$

$$3.5\times 10^{10}$$

Solution

Question 2

$$3\times 10^{-8}C$$

$$4\times 10^{-8}C$$

$$5\times 10^{-8}C$$

$$6\times 10^{-8}C$$

Solution

Question 3

Will increase

Will decrease

Will remain unchanged

Will increase if $$Q_{3}$$ is of the same sign as $$Q_{1}$$ and will decrease if $$Q_{3}$$ is of opposite sign

Solution

Question 4

$$9\times 10^3N$$

$$9\times 10^{-3}N$$

$$1.1\times 10^{-4}N$$

$$10^{-6}N$$

Solution

Question 5

Exist everywhere

Exist only in the immediate vicinity of electric charges

Exist only when both positive and negative charges are near one another

Are imaginary

Solution

Question 6

$$60 N$$

$$27 N$$

$$90 N$$

$$120 N$$

Solution

Question 7

$$d/\sqrt 2$$

$$d/2$$

$$\sqrt 2d$$

2 d

Solution

Question 8

2 F

4 F

8 F

16 F

Solution

Question 9

$$d\sqrt { 2 } $$

$$\dfrac { d }{ \sqrt { 2 } } $$

$$\dfrac { \sqrt { 2 } }{ d } $$

$$2d$$

Solution

Question 10

$${ 10 }^{ 16 }$$

$${ 10 }^{ 18 }$$

$${ 10 }^{ 20 }$$

$${ 10 }^{ 22 }$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.