Electric Charges and Fields Test

Result

Electric Charges and Fields Test - 36

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

A cylinder of radius $$R$$ and length $$l$$ is placed in a uniform electric field $$E$$ parallel to the axis of the cylinder. The total flux over the curved surface of the cylinder is :

A
Zero

B
$$\pi R^{2}E$$

C
$$2\pi R^{2}E$$

D
$$E/\pi R^{2}$$

Solution

For the curved surface,
At any point on the curved surface, $$E$$ and area vector are perpendicular to each other.
$$\quad \theta ={ 90 }^{ o }$$
$$\therefore \phi =E\quad ds\cos { { 90 }^{ o } } =0$$
Question 2
1 / -0

Identify the wrong statement in the following. Coulomb's law correctly describes the electric force that

A
binds the electrons of an atom to its nucleus

B
binds the protons and neutrons in the nucleus of an atom

C
binds atoms together to form molecules

D
binds atoms and molecules together to form solids

Solution

Nuclear force binds the protons and neutrons in the nucleus of an atom.
Question 3
1 / -0

Identify the correct statement about the charges $$q_1$$ and $$q_2$$ :

A
$$q_1$$ and $$q_2$$, both are positive

B
$$q_1$$ and $$q_2$$, both are negative

C
$$q_1$$ is positive and $$q_2$$ is negative

D
$$q_2$$ is positive and $$q_1$$ is negative

Solution

The filed of lines is outward from the positive charge and inward to the negative charge.
Here the field of lines is towards the charges $$q_1$$ and $$q_2$$. Thus, both charges are negative.
Question 4
1 / -0

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with another whose net charge is + 4Q. We move the +Q and +4Q charges to be 3 times as far apart as they were. What is the magnitude of the force on the +4Q charge ?

A
F

B
4F

C
4F/3

D
4F/9

E
F/3

Solution

Initially
$$q_1=q_2=Q$$ and $$r=R$$

Force was $$F=\dfrac{q_1q_2}{4\pi\epsilon_o r^2}=\dfrac{Q^2}{4\pi\epsilon_o R^2}$$

Thenafter:-
$$q_1=Q, q_2=4Q$$ and $$r=3R$$

Force on charge 4Q, $$F'=\dfrac{Q.4Q}{4\pi\epsilon_o (3R)^2}=\dfrac{4Q^2}{4\pi\epsilon_o (9R^2)}$$

$$\implies F'=\dfrac{4}{9}F$$

Answer-(D)
Question 5
1 / -0

A thin insulator rod is placed between two unlike point charges $$ +q_1 $$ and $$ q_2 $$ . For this situation tick the correct alternative (s)

A
The total force acting on charge $$ +q_1 $$ will increase

B
The total force acting on charge $$ -q_2 $$ will increase

C
The total force acting on charge $$ -q_2 $$ will decrease

D
The force acting on charge $$ +q_1 $$ due to $$ -q_2 $$ will remain same

Solution

Without rod the force between two charges is $$F=\dfrac{q_1(-q_2)}{4\pi \epsilon_0 r} $$ where $$\epsilon_0=$$ permittivity of free space
When rod inserted between charges , the force $$F'=\dfrac{q_1(-q_2)}{4\pi\epsilon}$$ where $$\epsilon=$$ permittivity of insulator medium.
Since $$\epsilon_0>\epsilon$$ so $$F'>F$$
According to Coulomb's law the force of attraction on both charges will equal.
Question 6
1 / -0

A particle A having a charge of $$2.0 \times 10^{-6}$$C and a mass of 100 g is fixed at the bottom of a smooth inclined plane of inclination $$30^{\circ}$$. Where should another particle B having same charge and mass, be placed on the inclined plane so that B may remain in equilibrium?(Take g= $$10 m/s^2$$)

A
8 cm from the bottom

B
13 cm from the bottom

C
21 cm from the bottom

D
27 cm from the bottom

Solution

$$\textbf{Step 1: Draw the free body diagram and Resolving forces[ Ref. Fig.]} $$
$$F_{e}$$ is the Electrostatic force experienced between particle A and B.
The weight $$mg$$ of the particle can be resolved in two components as shown in the figure:
Along the incline plane: $$mg\ sin\theta$$
Perpendicular to the incline plane: $$mg\ cos\theta$$
where $$\theta = 30^o$$

Let particle B be placed at a distance $$x$$ from particle A.

$$\textbf{Step 2: Newton's Second Law} $$
For particle B to remain in equilibrium the electrostatic force between the particles will be balanced by the component of the weight of particle B along incline direction i.e. acceleration $$=0$$

So, Applying Newton's Second Law along incline direction(Positive Upwards):
$$\Sigma F= ma$$
$$F_e- mg\sin 30^{\circ}=0$$
$$\Rightarrow\ \ F_{e} = mg \sin 30^{\circ} $$
$$\Rightarrow\ \ \dfrac{1}{4 \pi \epsilon_0} \dfrac{q_1 q_2}{x^2} = \dfrac{mg}{2} $$ $$....(1)$$

$$\textbf{Step 3: Substituting the values } $$
Equation $$(1)$$
$$\Rightarrow \dfrac{9 \times 10^{9} \times 2 \times 10^{-6} C\times 2 \times 10^{-6}C}{x^2} = \dfrac{0.1\ kg \times 10m/s^2}{2} $$
$$\Rightarrow \dfrac{36 \times 10^{-3}}{x^2} = \dfrac{1}{2} $$
$$\Rightarrow x^2 = 72 \times 10^{-3}\ m $$
$$\Rightarrow x = 26.8 \times 10^{-2}\ m = 27\ cm$$

Hence, Option D is correct.
Question 7
1 / -0

The electric field in a region of space is given by $$\vec{E} = 5\widehat{i} + 2\widehat{j} \ N/C$$. The flux of $$\vec{E}$$ due to this field through an area $$1m^2$$ lying in the y-z plane, in SI units, is :

A
$$5$$

B
$$10$$

C
$$2$$

D
$$5 \sqrt{29}$$

Solution

The flux , $$\phi=\vec{E}.\hat{n}da=(5\hat{i}+2\hat{j}).\hat{i}(1)=5 $$ as area y-z plane so normal vector is along x direction.
Question 8
1 / -0

In the diagram shown, the charge $$+Q$$ is fixed. Another charge $$+2q$$ and mass $$M$$ is projected from a distance $$R$$ from the fixed charge. Minimum separation between the two charges if the velocity becomes $$\displaystyle \frac{1}{\sqrt 3}$$ times of the projected velocity, at this moment is (Assume gravity to be absent):

A
$$\displaystyle \frac{\sqrt 3}{2}R$$

B
$$\displaystyle \frac{1}{\sqrt 3}R$$

C
$$\displaystyle \frac{1}{2}$$

D
$$None\ of\ these$$

Solution

The only force on the charges is electrostatic force, which is an internal force on this two charge system(+Q applies force on +2q and vice-versa). So conserving linear momentum is out of option, but angular momentum can be conserved along an axis passing through the +Q charge(As torque along this axis is zero because force is passing through the axis itself all the time). Also, at minimum distance, velocity of approach must be zero, i.e. velocity must be perpendicular to the line joining the charges.Hence:

$$\underset{L_{i}}{\rightarrow}\; =\; \underset{L_{f}}{\rightarrow}$$
$$\Rightarrow MVRSin(30^{\circ})\; =\; M(V/\sqrt{3})r$$
$$\Rightarrow r\; =\; \sqrt{3}/2\times R.$$
Question 9
1 / -0

Two charged particles are placed at a distance $$1.0 cm$$ apart. What is the minimum possible magnitude of the electric force acting on each charge?

A
$$23 \times 10^{-24} N$$

B
$$2.3 \times 10^{-24} N$$

C
$$2.3 \times 10^{-23} N$$

D
$$0$$

Solution

Electrostatic repulsion $$=F=\dfrac{1}{4 \pi \epsilon_o } \dfrac{q_1 q_2}{(.01)^2 } $$

force will be minimum when $$q_1=q_2 =$$ minimum unit of charge $$= \epsilon_o =1.6 \times 10^{-19} C $$

minimum force $$= 2.3 \times 10^{-24} N $$
Question 10
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
$$\displaystyle \frac{1}{16}$$

Solution

By Coulomb's law, the force between two charges is $$F=k \dfrac{q_1q_2}{r^2}$$ where r be the separation between charges.
Now, $$F'=k \dfrac{(2q_1)(2q_2)}{(r/2)^2}=16k\dfrac{q_1q_2}{r^2}=16F$$
Thus, $$n=16$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Electric Charges and Fields Test - 36

Result

Electric Charges and Fields Test - 36

Score

-

Rank

-

SHARING IS CARING

Question 1

Zero

$$\pi R^{2}E$$

$$2\pi R^{2}E$$

$$E/\pi R^{2}$$

Solution

Question 2

binds the electrons of an atom to its nucleus

binds the protons and neutrons in the nucleus of an atom

binds atoms together to form molecules

binds atoms and molecules together to form solids

Solution

Question 3

$$q_1$$ and $$q_2$$, both are positive

$$q_1$$ and $$q_2$$, both are negative

$$q_1$$ is positive and $$q_2$$ is negative

$$q_2$$ is positive and $$q_1$$ is negative

Solution

Question 4

F

4F

4F/3

4F/9

F/3

Solution

Question 5

The total force acting on charge $$ +q_1 $$ will increase

The total force acting on charge $$ -q_2 $$ will increase

The total force acting on charge $$ -q_2 $$ will decrease

The force acting on charge $$ +q_1 $$ due to $$ -q_2 $$ will remain same

Solution

Question 6

8 cm from the bottom

13 cm from the bottom

21 cm from the bottom

27 cm from the bottom

Solution

Question 7

$$5$$

$$10$$

$$2$$

$$5 \sqrt{29}$$

Solution

Question 8

$$\displaystyle \frac{\sqrt 3}{2}R$$

$$\displaystyle \frac{1}{\sqrt 3}R$$

$$\displaystyle \frac{1}{2}$$

$$None\ of\ these$$

Solution

Question 9

$$23 \times 10^{-24} N$$

$$2.3 \times 10^{-24} N$$

$$2.3 \times 10^{-23} N$$

$$0$$

Solution

Question 10

1

4

16

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

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.