Class 12 Physics Chapter 1 Electric Charges and Fields MCQs

Class 12 Physics Chapter 1 Electric Charges and Fields MCQs are provided here with answers. These questions are designed as per the latest CBSE syllabus and NCERT curriculum. Solving these chapter-wise MCQs will help students to score good marks in the final exam. Electric Charges and Fields Class 12 Physics MCQs are prepared for a better understanding of the concept. It allows students to test their knowledge and answering skills in the given time frame.

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MCQs on Class 12 Chapter 1 Electric Charges and Fields

Check the multiple-choice questions for the 12th Class Physics Electric Charges and Fields Chapter. Each MCQ will have four options here, out of which only one is correct. Students have to pick the correct option and check the answer provided here.

1. If the sizes of charged bodies are very small compared to the distances between them, we treat them as ————–.

  1. Zero charges
  2. Point charges
  3. Single charge
  4. No charges

Answer: (b) Point charges

Explanation: If the sizes of charged bodies are very small compared to the distances between them, we treat them as point charges.

2. The force per unit charge is known as ————

  1. Electric current
  2. Electric potential
  3. Electric field
  4. Electric space

Answer: (c) Electric field

Explanation: The force per unit charge is known as the electric field.

3. State true or false: The total charge of the isolated system is NOT conserved.

  1. True (b) False

Answer: (b) False

Explanation: As per the conservation of charges, it is said that the total charge of the isolated system is always conserved.

4. What is the dielectric constant of a metal?

  1. -1
  2. 0
  3. 1
  4. Infinite

Answer: (d) Infinite

Explanation: The dielectric constant of metals is infinite. The dielectric constant of metal is infinite, as the net electric field inside the metal is zero.

5. If the charge of 1 C is placed at a distance of 1 m from another charge of the same magnitude in a vacuum, it experiences an electrical force repulsion of magnitude ———-

  1. \(\begin{array}{l}9*10^{-9} N\end{array} \)
  2. \(\begin{array}{l}9*10^{9} N\end{array} \)
  3. \(\begin{array}{l}10*10^{9} N\end{array} \)
  4. \(\begin{array}{l}10*10^{-9} N\end{array} \)

Answer: (b)

\(\begin{array}{l}9*10^{9} N\end{array} \)

Explanation: If the charge of 1 C is placed at a distance of 1 m from another charge of the same magnitude in a vacuum, it experiences an electrical force repulsion of magnitude

\(\begin{array}{l}9*10^{9} N\end{array} \)
.

6. Quantisation of charge indicates that

  1. Charge, which is a fraction of charge on an electron, is not possible
  2. A charge cannot be destroyed
  3. Charge exists on particles
  4. There exists a minimum permissible charge on a particle

Answer: (a) Charge, which is a fraction of charge on an electron, is not possible

Explanation: Quantisation of charge indicates that

7. The property which differentiates two kinds of charges is called ————–

  1. Equality of charge
  2. Polarity of charge
  3. Fraction of charge
  4. None of the option

Answer: (b) Polarity of charge

Explanation: The property which differentiates two kinds of charges is called polarity of charge.

8. ————— gives the information on field strength, direction, and nature of the charge.

  1. Electric current
  2. Electric flux
  3. Electric field
  4. Electric potential

Answer: (c) Electric field

Explanation: Electric field gives the information on field strength, direction, and nature of the charge.

9.

\(\begin{array}{l}F=\frac{k.(q_{1}*q_{2})}{r^{2}}\end{array} \)
. This is given by which law?

  1. Faraday’s law
  2. Newton’s law
  3. Coulomb’s law
  4. Fleming’s law

Answer: (c) Coulomb’s law

Explanation:

\(\begin{array}{l}F=\frac{k.(q_{1}*q_{2})}{r^{2}}\end{array} \)
is given by Coulomb’s Law.

10. What happens when a glass rod is rubbed with silk?

  1. gains protons from silk
  2. gains electrons from silk
  3. gives electrons to silk
  4. gives protons to silk

Answer: (a) gains protons from silk.

Explanation: Excess electrons are transferred from glass to silk when a glass rod is rubbed with silk. Hence, the glass rod becomes positive, and silk becomes negative.

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