A 3 uF capacitor is charged to a potential of 300 V
and 2 pF capacitor is charged to 200 V. The
capacitors are then connected in parallel with plates
of opposite polarities joined together. What amount
of charge will flow, when the plates are so
connected?
(1) 1300 C
(2) 800 pc
(3) 600 C
(4) 300
Answers
The amount of charge flow when the plates are so connected will 600 μC.
Explanation:
=> Acoording to the question, a 3 μF capacitor is charged to a potential of 300 V.
Thus, charge q₁ = c*v = 3 * 300 * 10⁻⁶ = 900 μC
=> Similarly, 2 pF capacitor is charged to 200 V.
∴ q₂ = c*v = 2 * 200 * 10⁻⁶ = 400μC
=> Now, the capacitors are connected in parallel with plates of opposite polarities joined together then:
q₁' = C₁ (q₁ - q₂) / C₁ + C₂
= 3 (900 - 400) / 3 + 2
= 3 * 500 / 5
= 300 μC
q₂' = C₂ (q₁-q₂) / C₁ + C₂
= 2 (900 - 400) / 3 + 2
= 2 * 500 / 5
= 200 μC
=> Amount of charge flow
= q₁ - q₁'
= 900 - 300
= 600 μC
Thus, the amount of charge flow when the plates are so connected will 600 μC.
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