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A parallel plate capacitor of capacitance 5μF and plate separation 6 cm is connected to a 1 v battery and charged. A dielectric of dielectric constant 4 and thickness 4 cm is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is
  • a)
    2μc
  • b)
    3μc
  • c)
    5μc
  • d)
    10μc
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A parallel plate capacitor of capacitance 5μF and plate separation ...
as Q = cv
initially c = 5μF, v = 1v
∴ Q = 5 × 10–6 c = 5μc
when dielectric is introduced, let capacitance be c'
∴ c' = [(A∈0)/{d – t + (t/k)}]
here k = 4, t = 4 cm, d = 6 cm
∴ c' = [{(A∈0)/d}/{{d – t + (t/k)}/d}] ------ dividing by d
As [(A∈0)/d] = initially capacitance = c = 5μF
∴ c' = [c/{{6 – 4 + (4/4)}/6}] = [c/(3/6)] = 2c = 10μF
∴ Q' = c'v = 10 × 10–6 × 1 = 10μc
∴ addition charge flowing = Q' – Q = 10 – 5 = 5μc
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Most Upvoted Answer
A parallel plate capacitor of capacitance 5μF and plate separation ...
Microfarads has a charge of 10 microcoulombs stored on its plates. What is the potential difference between the plates?

We can use the formula for capacitance to find the potential difference:

C = Q/V

where C is the capacitance, Q is the charge, and V is the potential difference.

Rearranging the formula, we get:

V = Q/C

Substituting the given values, we get:

V = 10 microcoulombs / 5 microfarads

V = 2 volts

Therefore, the potential difference between the plates of the capacitor is 2 volts.
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