Concepts Of Physics MCQ Edition [Volume 2]PhysicsCapacitors
A parallel-plate capacitor with a capacitance of 1.2 imes 10⁻³ ext F has a charge of +2.0 imes 10⁻⁸ ext C deposited on its positive plate and -1.0 imes 10⁻⁸ ext C on its negative plate. Determine the potential difference that develops between the plates.
Options
- A8.33 10⁻⁶ V
- B1.67 10⁻⁵ V
- C1.25 10⁻⁵ V
- D2.50 10⁻⁵ V
Correct answer
C. 1.25 10⁻⁵ V
Step-by-step solution
When charges Q₁ and Q₂ are given to the two plates of a parallel-plate capacitor, the charge that appears on the inner facing surfaces is q = Q₁ - Q₂ 2 . Given Q₁ = 2.0 10⁻⁸ C and Q₂ = -1.0 10⁻⁸ C . Substituting the values: q = 2.0 10⁻⁸ - (-1.0 10⁻⁸) 2 q = 3.0 10⁻⁸ 2 = 1.5 10⁻⁸ C The potential difference V developed between the plates is given by V = q C . Given C = 1.2 10⁻³ F . V = 1.5 10⁻⁸ 1.2 10⁻³ V = 1.25 10⁻⁵ V