Electric Charges and Fields MCQ Test DPP 01

Electric Charges and Fields MCQ Test

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Q1.

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

Explanation: When a soap bubble is charged negatively, the excess electrons repel each other, which tends to push the surface outward. This increases the radius of the bubble.

Q2.

A body can be negatively charged by

Explanation: Negatively charging a body occurs when extra electrons are added to the body. Removing electrons would instead lead to a positive charge.

Q3.

The minimum possible charge on an object is \( 1.6 \times 10^{-19} \) coulomb.

Explanation: The basic unit of charge is the electron charge, \( 1.6 \times 10^{-19} \) coulomb. No charge can be smaller than an integral multiple of this charge.

Q4.

An attractive force between two neutrons is due to

Explanation: Neutrons are electrically neutral so they do not participate in electrostatic forces. Their attraction is explained by gravitational forces (very weak) and strong nuclear forces that bind nuclei.

Q5.

Two particles of equal mass \( m \) and charge \( q \) are placed at a distance of 16 cm. They do not experience any force. The value of \( \frac{q}{m} \) is:

Detailed Step-by-Step Explanation:

1. Write the Expressions for the Forces:
   • Coulomb’s force (repulsive): \[ F_e = \frac{1}{4\pi\epsilon_0}\frac{q^2}{l^2} \]
   • Gravitational force (attractive): \[ F_g = G\frac{m^2}{l^2} \]
2. Set the Forces Equal:
   Since the particles experience no net force, we have: \[ \frac{1}{4\pi\epsilon_0}\frac{q^2}{l^2} = G\frac{m^2}{l^2} \]
3. Simplify the Equation:
   Cancel the \( l^2 \) factor from both sides: \[ \frac{q^2}{4\pi\epsilon_0} = G\,m^2 \]
4. Solve for \( \frac{q}{m} \):
   Rearranging gives: \[ \frac{q^2}{m^2} = 4\pi\epsilon_0\,G \quad \Longrightarrow \quad \frac{q}{m} = \sqrt{4\pi\epsilon_0\,G} \]
5. Numerical Evaluation (Optional):
   Using the known constants:
   • \( \epsilon_0 \approx 8.85 \times 10^{-12} \) F/m
   • \( G \approx 6.67 \times 10^{-11} \) N·m²/kg²
   • \( 4\pi \approx 12.57 \)
   Multiply inside the square root: \[ 4\pi\epsilon_0\,G \approx 12.57 \times (8.85 \times 10^{-12}) \times (6.67 \times 10^{-11}) \approx 7.42 \times 10^{-21} \] Taking the square root: \[ \frac{q}{m} \approx \sqrt{7.42 \times 10^{-21}} \approx 8.61 \times 10^{-11}\; \text{C/kg} \]
6. Conclusion:
   The correct expression for \( \frac{q}{m} \) is \( \sqrt{4\pi\epsilon_0\,G} \), corresponding to Option (D).

Q6.

When the distance between the charged particles is halved, the force between them becomes

Explanation: According to Coulomb's law, the force between two charges is inversely proportional to the square of the distance between them. Halving the distance increases the force by a factor of \( 2^2 = 4 \).

Q7.

Number of electrons in one coulomb of charge will be approximately

Explanation: One coulomb of charge contains approximately \( \frac{1}{1.6 \times 10^{-19}} \) electrons, which is about \( 6.25 \times 10^{18} \). In this test, option (B) is taken as the correct approximate value.

Q8.

The electric charge in uniform motion produces

Explanation: A moving charge produces both an electric field (from the charge itself) and a magnetic field due to its motion. Hence, the correct answer is (C).

Q9.

Identify the wrong statement.

Explanation: Charge is a scalar quantity; stating that “charge is a vector quantity” is incorrect.

Q10.

If a charge on the body is \(-1\,\text{nC}\), then how many number of excess electrons are present on the body?

Detailed Step-by-Step Explanation:

1. Understanding the Problem:
   The question asks for the number of excess electrons corresponding to a net charge of \(-1\,\text{nC}\). Although the charge is negative, we use its magnitude to find the number of electrons.
2. Identify the Fundamental Charge:
   The charge of one electron is approximately \[ e = 1.6 \times 10^{-19}\,\text{C}. \]
3. Convert the Given Charge:
   A charge of \(-1\,\text{nC}\) is equivalent to \[ 1\,\text{nC} = 1 \times 10^{-9}\,\text{C}. \]
4. Calculate the Number of Electrons:
   The number of electrons is given by: \[ \text{Number of electrons} = \frac{1 \times 10^{-9}\,\text{C}}{1.6 \times 10^{-19}\,\text{C/electron}} \approx 6.25 \times 10^{9}\,\text{electrons}. \]
5. Conclusion:
   The body has approximately \(6.25 \times 10^{9}\) excess electrons. Hence, option (B) is the correct answer.

Q11.

A cylindrical conductor is placed near another positively charged conductor. The net charge acquired by the cylindrical conductor will be

Explanation: When an uncharged conductor is brought near a charged conductor, charge redistribution (induction) occurs but the net charge remains zero.