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प्रश्न
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3m2 and the separation between the plates is 3 mm.
- Calculate the capacitance of the capacitor.
- If this capacitor is connected to 100 V supply, what would be the charge on each plate?
- How would charge on the plates be affected, if a 3 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected?
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उत्तर
Here, A = 6 × 10–3m2, d = 3 mm = 3 × 10–3m
- Capacitance, C = `(∈_0A)/d`
= `((8.85 xx 10^-12 xx 6 xx 10^-3))/(3 xx 10^-3)`
= 17.7 × 10–12F - Charge, Q = CV
= 17.7 × 10–12 × 100
= 17.7 × 10–10C - New charge Q' = KQ
= 6 × 17.7 × 10–10
= 1.062 × 10–8C
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(i) electric field between the plates
(ii) capacitance, and
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A parallel-plate capacitor has plate area 20 cm2, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy of the capacitor 8.9 μs after the connections are made.
A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations:

- Key K is kept closed and plates of capacitors are moved apart using insulating handle.
- Key K is opened and plates of capacitors are moved apart using insulating handle.
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- In A: Q remains same but C changes.
- In B: V remains same but C changes.
- In A: V remains same and hence Q changes.
- In B: Q remains same and hence V changes.
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