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

12th Grade > Physics

ELECTRIC CHARGES FIELDS AND GAUSS LAW MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1. Electric charge is uniformly distributed along a long straight wire of radius 1mm. The charge per cm length of the wire is Q coulomb. Another cylindrical surface of radius 50 cm and length 1m symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is

Qϵ0
100Qϵ0
10Q(πϵ0)
100Q(πϵ0)

Discuss Question

Answer: Option B. -> 100Qϵ0
:
B
Charge enclosed by cylindrical surface (length 100 cm) is

Q_{e x c} = 100 Q .

By applying Gauss's law

ϕ = \frac{1}{ϵ_{0}} (Q_{e γ α}) = \frac{1}{ϵ_{0}} (100 Q)

Question 2. The distance between charges

5 \times 10^{- 11}

C and

- 2.7 \times 10^{- 11}

C is

0.2 m

. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is

0.44 m
0.65 m
0.556 m
0.350 m
If assertion is false but reason is true.

Discuss Question

Answer: Option C. -> 0.556 m
:
C
If two opposite charges are separated by a certain distance, then for it's equalibrium a third
charge should be kept outside and near the charge which is smaller in magnitude.
Here, suppose third charge

q

is placed at a distance

x

from

- 2.7 \times 10^{- 11} C

then for it's
equalibrium

| F_{1} | = | F_{2} |

The Distance Between Charges 5×10−11C And −2.7×10−...

\Rightarrow \frac{K Q_{1} q}{(x + 0.2)^{2}} = \frac{K Q_{2} q}{x^{2}} \Rightarrow x = 0.556 m

(H e r e K = \frac{1}{4 π ϵ_{0}} a n d Q_{1} = 5 \times 10^{- 11} C, Q_{2} = - 2.7 \times 10^{- 11} C)

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}}

Two Similar Spheres Having +q And -q Charge Are Kept At A Ce...

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. 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 5. 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
−q2
4q
If assertion is false but reason is true.

Discuss Question

Answer: Option A. -> -q
:
A
The force between 4qand q;

F_{1} = \frac{1}{4 π ϵ_{0}} . \frac{4 q \times q}{l^{2}}

The force between Qand q;

F_{2} = \frac{1}{4 π ϵ_{0}} . \frac{Q \times q}{(\frac{l}{2})^{2}}

We want

F_{1} + F_{2} = 0

\frac{4 q^{2}}{l^{2}} = - \frac{4 Q q}{l^{2}} \Rightarrow Q = - q

Question 6. A soap bubble is given a negative charge, then its radius

Decreases
Increases
Remains unchanged
Nothing can be predicted as information is insufficient

Discuss Question

Answer: Option B. -> Increases
:
B
Due to mutual repulsion of charges distributed on the surface of bubble.

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√2
10√2
20

Discuss Question

Answer: Option C. -> 10√2
:
C
Body moves along the parabolic path.

There Is A Uniform Electric Field Of Strength 103 V/m Along...

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. 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 9. A conductor has

14.4 \times 10^{- 19}

coulombs positive charge. The conductor has (Charge on electron

1.6 \times 10 - 19

coulombs )

9 electrons in excess
27 electrons in short
27 electrons in excess
9 electrons in short
If assertion is false but reason is true.

Discuss Question

Answer: Option D. -> 9 electrons in short
:
D
Positive charge shows the deficiency of electrons. number of electrons

= \frac{14.4 \times 10^{- 19}}{1.6 \times 10^{- 19}} = 9

Question 10. 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 toPandSand due toQandTadd to zero, while due toUandRwill be added up.