Static Electricity MCQ

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Static electricity is caused due to the imbalances of positive and negative charges. These charges produced would not be able to move until an external electric current is applied. Static electricity is the reason why pieces of paper stick to a balloon’s surface when it is rubbed on a woollen cloth. Static electricity is seen in many daily life applications. Lightning is one major example.

There are two types of charges.

Positive Charge: When an atom loses electrons, it acquired positive charges.
Negative Charge: When an atom gains electrons, it acquired negative charges.

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MCQs on Static Electricity

Ques 1. Electric current originates from which part of an atom?

Nucleus
Entire atom acting as a unit
Positively charged protons
Negatively charged electrons

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Answer: d

Explanation: Electric current is basically the movement of electrons. Thus it originates from the negatively charged electrons.

Ques 2. Which of the following is true of electrical forces?

Like charges attract, unlike charges repel.
Electrical forces are produced by electrical charges.
Positive and negative charges can combine to produce the third type.
Electrical forces are weaker than gravitational forces.

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Answer: b

Explanation: The movement of electrical charges produces electrical forces.

Ques 3. When a person combs her hair, static electricity is sometimes generated by what process?

Friction between the comb and hair transfers electrons.
Induction between the comb and hair.
Deduction between the comb and hair.
Contact between the comb and hair results in a charge.

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Answer: a

Explanation: When a person combs their dry hair, friction occurs between the dry hair and the teeth of the comb. This results in the exchange of charges which constitutes electrostatic force.

Ques 4. Attraction is the actual proof of an object being charged.

True
False

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Answer: b

Explanation: There are even possibilities that an uncharged object can exhibit attraction by another charged object. So being attracted does not necessarily mean that the body is charged.

Ques 5. If an object is positively charged, theoretically the mass of the object ______

Increases slightly by a factor of 9.11*10^-31 kg
Decreases slightly by a factor of 9.11*10^-31 kg
Remains the same
May increase or decrease

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Answer: b

Explanation: It is a known fact that if an object is positively charged it loses some of its electrons. The mass of an electron is 9.11x10^-31 kg. So, if a positively charged body loses ‘n’ number of electrons, its mass decrease by the amount nx9.11x10^-31kg.

Ques 6. Why is gold used in the Gold-leaf electroscope?

Gold is easily available in nature
Gold is malleable
Gold is conducting in nature
Gold is cheap

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Answer: b

Explanation: Gold is a very malleable metal. This means very thin and light sheets can be formed from gold simply by hammering or rolling and hence the deflection of the light gold plates increases.

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Ques 7. What happens to the plates of the apparatus if we measure alternating charge using a Gold-leaf oscilloscope?

It doesn’t diverge at all
It diverges momentarily
The plates give a proper divergence
The degree of divergence increases and decreases repeatedly

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Answer: c

Explanation: The divergence of the plates of the Gold-leaf oscilloscope depends only on the presence of a charge, not on the quality of charge i.e. positive or negative. So, if the charge changes from positive to negative and vice versa the degree of divergence of the plates remains the same.

Ques 8. Why can’t we induce static electricity in a brass road by friction if we hold it in out hand?

Static electricity is induced only on insulators while brass is conducting
The induced static charge will flow through our body to the ground
The static charge was not produced at all
It is not possible to induce static electricity by friction

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Answer: b

Explanation: The human body is a good conductor and hence the charge produced by friction flows through our body to the ground. This can be overcome by holding the brass rod with some insulator like wood or plastic.

Ques 9. A gold-leaf electroscope is used for?

Only to detect the presence of charge
To detect the presence of charge as well as its nature (positive or negative)
To measure the surface charge density
To measure current

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Answer: b

Explanation: A gold leaf electroscope is primarily used to detect the nature of charges. If the apparatus is initially in contact with some positively charged body and then comes in contact with another charged body and if the divergence increases we can deduce that the second body is also positively charged. If the divergence decreases then the second body must be negatively charged.

Ques 10. 1 Coulomb = ________ Electro Static Unit.

3*10⁹
3*10⁸
1.602*10^-19
2*10⁹

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Answer: a

Explanation: 1 Coulomb is equal to 3*10⁹ Electro Static Unit of charge. Coulomb is the unit of charge in the SI system and esu is the unit of charge in the CGS system.

Ques 11. Number of electrons in 1 Coulomb charge is _________

6.25 * 10²¹
6.25 * 10²⁰
6.25 * 10²⁵
6.25 * 10²³

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Answer: a

Explanation: Electron has a charge of 1.602*10^-19 Coulomb. So, the number of electrons in 1 Coulomb of charge is 1/1.602∗10^-19 = 6.25*10²¹.

Ques 12. What happens if two equally and oppositely charged spheres touch each other?

Attract each other with the same amount of force
Repel each other
Attract each other with less amount of force
Don’t attract or repel

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Answer: d

Explanation: Since the two spheres are equally and oppositely charged, the charges neutralize each other and the spheres neither attract nor repel each other.

Ques 13. Which among the following cannot be the charge of a charged body?

4.8 * 10^-14 Coulomb
6.4 * 10^-15 Coulomb
5 * 10^-14Coulomb
3.2 * 10^-10 Coulomb

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Answer: c

Explanation: Charge of a body must be an integral multiple of the charge of an electron 1.602∗10^-19. Thus, 5*10^-14 is not an integral multiple of 1.602∗10^-19.

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Static Electricity MCQ

MCQs on Static Electricity

CBSE CLASS XII Related Questions

1.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

2.
The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

3.
The resistance of a wire at 25°C is 10.0 \( \Omega \). When heated to 125°C, its resistance becomes 10.5 \( \Omega \). Find (i) the temperature coefficient of resistance of the wire, and (ii) the resistance of the wire at 425°C.

4.
The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

5.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

6.
A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is:

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Static Electricity MCQ

MCQs on Static Electricity

CBSE CLASS XII Related Questions

1.Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

2.The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

3.The resistance of a wire at 25°C is 10.0 \( \Omega \). When heated to 125°C, its resistance becomes 10.5 \( \Omega \). Find (i) the temperature coefficient of resistance of the wire, and (ii) the resistance of the wire at 425°C.

4.The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

5.Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

6.A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

2.
The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

3.
The resistance of a wire at 25°C is 10.0 \( \Omega \). When heated to 125°C, its resistance becomes 10.5 \( \Omega \). Find (i) the temperature coefficient of resistance of the wire, and (ii) the resistance of the wire at 425°C.

4.
The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

5.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

6.
A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is: