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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
संबंधित प्रश्न
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Solve the following question.
A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.
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What is the capacitance of a parallel plate capacitor with a dielectric medium of dielectric constant \[K\]?
What is the effect of inserting a dielectric with \[K>1\] between the plates?
When a dielectric slab of dielectric constant \[K\] is inserted between the plates, what is the electric field \[E'\] due to polarisation of the dielectric?
