Electric Charges and Fields Test

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Electric Charges and Fields Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

A charged particle moves in an electric field from $A$ to $B$ and then $B$ to $A$ .

A
If $W_{AB} > W_{BA}$ , then the field is conservative

B
If $W_{AB} + W_{BA} = 0$ , then the field is conservative

C
If $W_{AB} > W_{BA} > 0$ , then the field is conservative

D
If $W_{AB} = W_{BA}$ , then the field is conservative

Solution
Question 2
1 / -0

A metallic particle having no net charge is placed near a finite metal plate carrying a positive charge. The electric force on the particle will be

A
towards the plate

B
away from the plate

C
parallel to the plate

D
zero

Solution

Since there is no charge on the metallic particle therefore, force experienced by the particle is zero.
Hebce option A is correct.
Question 3
1 / -0

Number of electric lines of force from $0.5\ C$ of positive charge in a dielectric medium of dielectric constant $10$ is

A
$5.65\times 10^{9}$

B
$1.13\times 10^{11}$

C
$9\times 10^{9}$

D
$8.85\times 10^{-12}$

Solution

Given data $Q=0.5C, k= 10$
We know the formula,
$=\dfrac {Q}{K\epsilon_{0}}$
$= \dfrac {0.5}{10\times 8.85\times 10^{-12}} = 5.65\times 10^{9}$ .
Question 4
1 / -0

If a body is charged by rubbing it, its weight

A
always decreases slightly

B
always increases slightly

C
may increase slightly or may decrease slightly

D
remains precisely the same

Solution

If a body is charged by rubbing it, then it may lose or gain electrons. Since electrons have a mass of $(9.1\times 10^{-31}kg).$ So, a slight weight may increase or decrease slightly.
Question 5
1 / -0

An electron is released from rest in a region of space with a non-zero electric field. Which of the following statements is true?

A
The electron will begin moving towards a region of higher potential.

B
The The electron will begin moving towards a region of lower potential.

C
The The electron will begin moving among a line of constant potential.

D
Nothing can be concluded unless the direction of the electric field is known.

Solution

Negative charge itself goes from low to high potential.
Question 6
1 / -0

Three charges of $q_{1} = 1\times 10^{-6}C, q_{2} = 2\times 10^{-6}C$ and $q_{3} = -3\times 10^{-6}C$ have been placed as shown. Then, the net electric flux will be maximum for the surface.

A
$S_{1}$

B
$S_{2}$

C
$S_{3}$

D
Same for all three

Solution

Given:
$q_1=1\times 10^{-6}\\q_2=2\times 10^{-6}\\q_3=-3\times 10^{-6}$
Electric flux $=\dfrac{\phi}{\epsilon_0}$
Charge inside $S_{1} = q_{1} + q_{2} = 3\times 10^{-6} C$
Charge inside $S_{2} = q_{2} + q_{3} = -1\times 10^{-6} C$
Charge inside $S_{3} = 0$
Charge inside $S_{1}$ is greatest. So, flux through $S_{1}$ is maximum.
Question 7
1 / -0

F is the force between two charges. If the distance between them is tripled, then the force between the charges will be :

A
F

B
$\dfrac{F}{3}$

C
$\dfrac{F}{9}$

D
$\dfrac{F}{27}$

Solution

$F=\dfrac{1}{4\pi \epsilon_0}\,\dfrac{q_1q_2}{r^2}$ and

$F'=\dfrac{1}{4\pi \epsilon_0}\,\dfrac{q_1q_2}{(3r)^2}$

$=\dfrac{1}{9}\Bigg(\dfrac{1}{4\pi \epsilon_0}\,\dfrac{q_1q_2}{r^2}\Bigg)=\dfrac{F}{9}$
Question 8
1 / -0

An attractive force of $9$ N acts between $+5$ C and $-5$ C at some distance. These charges are allowed to touch each other and are then again placed at their initial position. The force acting between them will be :

A
infinite

B
$9 \times 10^9$ N

C
$1$ N

D
zero

Solution

According to the question
$F=\dfrac{1}{4\pi \epsilon_0}\,\dfrac{(5 \times 10^{-6})\times(-5 \times 10^{-6})}{r^2}$
On touching them the charge on each will be zero.
$\therefore$ $F=\dfrac{1}{4\pi \epsilon_0}\,\dfrac{(0)}{r^2}=0$
Question 9
1 / -0

A very small ball has a mass of $5.00 \times 10^{-3} \,kg$ and
a charge of $4.00 \,mC$ . What magnitude electric field directed upward will balance the weight of the ball so that the ball is suspended motionless above the ground?

A
$8.21 \times 10^{2} N/C$

B
$1.22 \times 10^4 N/C$

C
$2.00 \times 10^{-2} N/C$

D
$5.11 \times 10^6 N/C$

E
$3.72 \times 10^3 N/C$

Solution

To balance the weight of the ball, the magnitude of the upward electric force must equal the magnitude of the downward gravitational force, or $qE =mg$ , which gives
$E=\dfrac{mg}{q}=\dfrac{(5.0\times10^{-3}(9.80 m/s^2)}{4.0\times 10^{-6} C}=1.2\times 10^4 N/C$
Question 10
1 / -0

Two point charges attract each other with an electric force of magnitude F. If the charge on one of the particles is reduced to one-third its original value and thedistance between the particles is doubled, what is the resulting magnitude of the electric force between them?

A
$\dfrac{1}{12}F$

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

C
$\dfrac{1}{6}F$

D
$\dfrac{3}{4}F$

E
$\dfrac{3}{2}F$

Solution

The magnitude of the electric force between charges $Q_1$ and $Q_2$ , separated by distance $r_i$ is $F=k_eQ_1Q_2/r^2$ . if changes are made so $Q_1\rightarrow Q_1/3$ and $r \rightarrow 2r$ , the magnitude of the new force $F'$ will be
$F'=k_e=\dfrac{(Q_1/3)Q_2}{(2r)^2}=\dfrac{1}{3(4)}k_e\dfrac{Q_1 Q_2}{r^2}=\dfrac{1}{12}k_e\dfrac{Q_1 Q_2}{r^2}=\dfrac{1}{12}F$

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Electric Charges and Fields Test - 63

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Electric Charges and Fields Test - 63

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Question 1

If WAB>WBAW_{AB} > W_{BA}WAB​>WBA​, then the field is conservative

If WAB+WBA=0W_{AB} + W_{BA} = 0WAB​+WBA​=0, then the field is conservative

If WAB>WBA>0W_{AB} > W_{BA} > 0WAB​>WBA​>0, then the field is conservative

If WAB=WBAW_{AB} = W_{BA}WAB​=WBA​, then the field is conservative

Solution

Question 2

towards the plate

away from the plate

parallel to the plate

zero

Solution

Question 3

5.65×1095.65\times 10^{9}5.65×109

1.13×10111.13\times 10^{11}1.13×1011

9×1099\times 10^{9}9×109

8.85×10−128.85\times 10^{-12}8.85×10−12

Solution

Question 4

always decreases slightly

always increases slightly

may increase slightly or may decrease slightly

remains precisely the same

Solution

Question 5

The electron will begin moving towards a region of higher potential.

The The electron will begin moving towards a region of lower potential.

The The electron will begin moving among a line of constant potential.

Nothing can be concluded unless the direction of the electric field is known.

Solution

Question 6

S1S_{1}S1​

S2S_{2}S2​

S3S_{3}S3​

Same for all three

Solution

Question 7

F

F3\dfrac{F}{3}3F​

F9\dfrac{F}{9}9F​

F27\dfrac{F}{27}27F​

Solution

Question 8

infinite

9×1099 \times 10^99×109 N

111 N

zero

Solution

Question 9

8.21×102N/C8.21 \times 10^{2} N/C8.21×102N/C

1.22×104N/C1.22 \times 10^4 N/C1.22×104N/C

2.00×10−2N/C2.00 \times 10^{-2} N/C2.00×10−2N/C

5.11×106N/C5.11 \times 10^6 N/C5.11×106N/C

3.72×103N/C3.72 \times 10^3 N/C3.72×103N/C

Solution

Question 10

112F\dfrac{1}{12}F121​F

13F\dfrac{1}{3}F31​F

16F\dfrac{1}{6}F61​F

34F\dfrac{3}{4}F43​F

32F\dfrac{3}{2}F23​F

Solution

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If $W_{AB} > W_{BA}$ , then the field is conservative

If $W_{AB} + W_{BA} = 0$ , then the field is conservative

If $W_{AB} > W_{BA} > 0$ , then the field is conservative

If $W_{AB} = W_{BA}$ , then the field is conservative

$5.65\times 10^{9}$

$1.13\times 10^{11}$

$9\times 10^{9}$

$8.85\times 10^{-12}$

$S_{1}$

$S_{2}$

$S_{3}$

$\dfrac{F}{3}$

$\dfrac{F}{9}$

$\dfrac{F}{27}$

$9 \times 10^9$ N

$1$ N

$8.21 \times 10^{2} N/C$

$1.22 \times 10^4 N/C$

$2.00 \times 10^{-2} N/C$

$5.11 \times 10^6 N/C$

$3.72 \times 10^3 N/C$

$\dfrac{1}{12}F$

$\dfrac{1}{3}F$

$\dfrac{1}{6}F$

$\dfrac{3}{4}F$

$\dfrac{3}{2}F$