Electric Fields, Capacitors, and Related Concepts – Step-by-Step Physics Guidance

Q4. The figure shows electric field lines arising from two small charged particles P and Q. Consider the following two statements:

• (i) The charge on P is smaller than the charge on Q.

• (ii) The electrostatic force on P is smaller than the force on Q.

Which is true?

Background

Topic: Electric Field Lines and Electrostatic Forces

This question tests your understanding of how electric field lines relate to the magnitude of charge and the forces experienced by charged particles.

Key Terms and Formulas

• Electric field lines: The number of lines emerging from a charge is proportional to the magnitude of the charge.

• Electrostatic force: The force between two charges is given by Coulomb's Law:

$F = \frac{k |q_1 q_2|}{r^2}$

• $F$ = force between charges

• $k$ = Coulomb's constant

• $q_1, q_2$ = charges

• $r$ = distance between charges

Step-by-Step Guidance

1. Observe the number of electric field lines coming from each charge. More lines indicate a larger charge.

   

2. Recall that the force experienced by a charge depends on both the magnitude of the charge and the electric field it is in.

3. Compare the statements: (i) relates to the charge magnitude, (ii) relates to the force experienced by each charge.

4. Think about how the electric field lines and Coulomb's Law help you determine which statements are true.

Try solving on your own before revealing the answer!

Final Answer: (a) Only (i) is true.

There are more field lines coming from Q than P, so Q has a larger charge. The force experienced by each is equal in magnitude (Newton's third law), so statement (ii) is false.

Q8. The figure shows two parallel-plate capacitors storing the same amount of charge Q. The plates of each capacitor are separated by the same distance, and their areas are identical. Capacitor 2 is filled with a dielectric material with dielectric constant κ (> 1). Consider the following statements:

• (i) The capacitance of Capacitor 1 is larger than that of Capacitor 2.

• (ii) The potential energy stored in Capacitor 1 is larger than that in Capacitor 2.

Which is true?

Background

Topic: Capacitance and Dielectrics

This question tests your understanding of how a dielectric affects the capacitance and energy storage of a parallel-plate capacitor.

Key Terms and Formulas

• Capacitance without dielectric: $C_1 = \frac{\varepsilon_0 A}{d}$

• Capacitance with dielectric: $C_2 = \kappa \frac{\varepsilon_0 A}{d}$

• Potential energy stored: $U = \frac{Q^2}{2C}$

• $\varepsilon_0$ = permittivity of free space

• $A$ = plate area

• $d$ = plate separation

• $\kappa$ = dielectric constant ($\kappa > 1$)

Step-by-Step Guidance

1. Compare the capacitance formulas for both capacitors. The dielectric increases capacitance.

   

2. Recall that both capacitors store the same charge $Q$.

3. Use the formula for potential energy to compare the energy stored in each capacitor.

4. Think about how increasing capacitance affects the stored energy for a fixed charge.

Try solving on your own before revealing the answer!

Final Answer: (b) Only (ii) is true.

Capacitor 2 has a larger capacitance due to the dielectric, so statement (i) is false. For the same charge, the energy stored is less in the capacitor with higher capacitance, so statement (ii) is true.