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 Force

This question tests your understanding of how electric field lines relate to the magnitude of charge and the force 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 around P and Q. More lines indicate a larger charge magnitude.

   

2. Recall that the force experienced by a charge in an electric field depends on both the charge and the strength of the field.

3. Compare the field lines: If Q has more lines than P, Q has a greater charge magnitude.

4. Think about how the force on each charge relates to the electric field and the charge itself.

Try solving on your own before revealing the answer!

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

Q has more field lines, so its charge is larger. The force on each charge depends on the other charge and the field, but the field lines only directly indicate charge magnitude, not force.

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: $C_1 = \varepsilon_0 \frac{A}{d}$

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

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

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

• $A$ = plate area, $d$ = plate separation, $\varepsilon_0$ = permittivity of free space

Step-by-Step Guidance

1. Compare the capacitance formulas: Adding a dielectric increases capacitance ($C_2 > C_1$).

   

2. Recall that for the same charge $Q$, the potential energy stored is inversely proportional to capacitance.

3. Analyze how the dielectric affects both capacitance and stored energy.

4. Think about which statements (i) and (ii) are correct based on the relationships above.

Try solving on your own before revealing the answer!

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

Capacitor 2 has greater capacitance due to the dielectric, so statement (i) is false. For the same charge, Capacitor 1 stores more energy because its capacitance is lower, making statement (ii) true.