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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?
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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
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
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.
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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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 isa)2μcb)3μcc)5μcd)10μcCorrect answer is option 'C'. Can you explain this answer?
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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 isa)2μcb)3μcc)5μcd)10μcCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 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 isa)2μcb)3μcc)5μcd)10μcCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 isa)2μcb)3μcc)5μcd)10μcCorrect answer is option 'C'. Can you explain this answer?.
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