Electric Charges Fields And Gauss Law(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

ELECTRIC CHARGES FIELDS AND GAUSS LAW MCQs

Total Questions : 30 | Page 1 of 3 pages

When air is replaced by a dielectric medium of constant k , the maximum force of attraction between two charges separated by a distance

Decreases k times
Remains unchanged
Increases k times
Increases k^-1 times

Discuss Question

Answer: Option A. -> Decreases k times
:
A

$F^{'} = \frac{1}{4 π ϵ_{0}} \frac{q_{1} q_{2}}{r^{2}} = \frac{F}{K}$
If F is the force in air, then F is less than F since $K > 1.$

Question 2.

Four charges are arranged at the corners of a square ABCD , as shown in the adjoining figure. The force on the charge kept at the centre O is

Zero
Along the diagonal AC
Along the diagonal BD
Perpendicular to side AB

Discuss Question

Answer: Option C. -> Along the diagonal BD
:
C

We put a unit positive charge at O. Resultant force due to the charge placed at A and C is zero and resultant charge due to B and D is towards D along the diagonal BD.

Question 3.

Two similar spheres having +q and -q charge are kept at a certain distance. F force acts between the two. If in the middle of two spheres, another similar sphere having +q charge is kept, then it experiences a force whose magnitude and direction are

Zero having no direction
8F towards +q charge
8F towards -q charge
4F towards +q charge

Discuss Question

Answer: Option C. -> 8F towards -q charge
:
C

Initially, force betwen A and C $F = K \frac{Q^{2}}{r^{2}}$

When a similar sphere B having charge +Q is kept at the mid point of line joining A and
C, then Net force on B is
$F_{n e t} = F_{A} + F_{C} = K \frac{Q^{2}}{(\frac{r}{2})} + \frac{K Q^{2}}{(\frac{r}{2})^{2}} = 8 \frac{K Q^{2}}{r^{2}} = 8 F .$

(Direction is shown in figure)

Question 4.

The electric field due to a charge at a distance of 3 m from it is 500 N/coulomb. The magnitude of the charge is $[\frac{1}{4 π ϵ_{0}} = 9 \times 10^{9} \frac{N - m^{2}}{c o u l o m b^{2}}]$

2.5 micro-coulomb
2.0 micro-coulomb
1.0 micro-coulomb
0.5 micro-coulomb

Discuss Question

Answer: Option D. -> 0.5 micro-coulomb
:
D

$E = 9 \times 10^{9} \times \frac{Q}{r^{2}} \Rightarrow 500 = 9 \times 10^{9} \times \frac{Q}{(3)^{2}} \Rightarrow Q = 0.5 μ C$

Question 5.

When a glass rod is rubbed with silk, it

Gains electrons from silk
Gives electrons to silk
Gains protons from silk
Gives protons to silk

Discuss Question

Answer: Option B. -> Gives electrons to silk
:
B

On rubbing glass rod with silk, excess electron transferred from glass to silk. So glass rod becomes positive and silk becomes negative.

Question 6.

Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon such that the electric field at O is double the electric field when only one positive charge of same magnitude is placed at R. Which of the following arrangements of charges is possible for P, Q, R, S, T and U respectively

+,-,+,-,-,+
+,-,+,-,+,-
+,+,-,+,-,-
-,+,+,-,+,-

Discuss Question

Answer: Option D. -> -,+,+,-,+,-
:
D

If the charges are arranged according to the option (d), the electric fields due to P and S and due to Q and T add to zero, while due to U and R will be added up.

Question 7.

There is a uniform electric field of strength $10^{3}$ V/m along y-axis. A body of mass 1g and charge $10^{- 6}$ C is projected into the field from origin along the positive x-axis with a velocity 10m/s. Its speed in m/s after 10s is (Neglect gravitation)

10
$5 \sqrt{2}$
$10 \sqrt{2}$
20

Discuss Question

Answer: Option C. ->

10 \sqrt{2}

:
C

Body moves along the parabolic path.

For vertical motion : By using $v = u + a t$
$\Rightarrow v_{y} = 0 + \frac{Q E}{m} . t = \frac{10^{- 6} \times 10^{3}}{10^{- 3}} \times 10 = 10 m / s e c$
For horizontal motion - it;s horizontal velocity remains the same i.e after the same i.e after 10 sec, horizontal
velocity of body $v_{x} = 10 m / s e c .$
Velocity aftey 10 sec $v = \sqrt{V_{x}^{2} + V_{y}^{2}} = 10 \sqrt{2} m / s e c$

Question 8.

An electron enters with its velocity in the direction of the uniform electric lines of force. Consider lines of force to be straight. Then

The path of the electron will be a circle
The path of the electron will be a parabola
The magnitude of velocity of the electron will first decrease and then increase
The magnitude of velocity of the electron will first increase and then decrease

Discuss Question

Answer: Option C. -> The magnitude of velocity of the electron will first decrease and then increase
:
C

As electric field applies the force on electron in the direction opposite to it’s motion, the velocity of the electron will decrease.

Question 9.

Assertion: The coulomb force is the dominating force in the universe.
Reason :The coulomb force is weaker than the gravitational force.

If both assertion and reason are true and the reason is the correct explanation of the assertion.
If both assertion and reason are true but reason is not the correct explanation of the assertion.
If assertion is true but reason is false.
If the assertion and reason both are false.
If assertion is false but reason is true.

Discuss Question

Answer: Option D. -> If the assertion and reason both are false.
:
D

Gravitational force is the dominating force in nature and not coulomb's force. Gravitational force is the weakest force. Also, Coulomb's force>> gravitational force.

Question 10.

Three charges 4q, Q and q are in a straight line in the position of 0, $\frac{l}{2}$ and l respectively. The resultant force on q will be zero, if Q =

-q
-2q
$- \frac{q}{2}$
4q
If assertion is false but reason is true.

Discuss Question

Answer: Option A. -> -q
:
A

The force between 4q and q; $F_{1} = \frac{1}{4 π ϵ_{0}} . \frac{4 q \times q}{l^{2}}$
The force between Q and q; $F_{2} = \frac{1}{4 π ϵ_{0}} . \frac{Q \times q}{(\frac{l}{2})^{2}}$
We want $F_{1} + F_{2} = 0$ or $\frac{4 q^{2}}{l^{2}} = - \frac{4 Q q}{l^{2}} \Rightarrow Q = - q$