Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) While the voltage supply remained connected. (b) After the supply was disconnected.

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(a) Dielectric constant of the mica sheet, k = 6 Initial capacitance, C = 1.771 × 10−11 F Supply voltage, V = 100 V Potential across the plates remains 100 V. (b) Dielectric constant, k = 6 Initial capacitance, C = 1.771 × 10−11 F If supply voltage is removed, then there will...
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(a) Dielectric constant of the mica sheet, k = 6 Initial capacitance, C = 1.771 × 10−11 F Supply voltage, V = 100 V Potential across the plates remains 100 V. (b) Dielectric constant, k = 6 Initial capacitance, C = 1.771 × 10−11 F If supply voltage is removed, then there will be no effect on the amount of charge in the plates. Charge = 1.771 × 10−9 C Potential across the plates is given by, read less
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Have been a faculty with pace, yukti and vidyalankar

When the power supply remains connected, V will not change. Power supply maintains voltage.When the power supply is disconnected, the charge will remain the same. The charge is trapped.
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Mechanical engineer with 2 years of teaching experience

We know, capacitance = dielectric constant × C₀C = KC₀ = 6 × 18pF = 108pF (a) if the voltage supply remained connected , then the potential difference across the capacitor will remain the same. e.g., V = 100V and hence, charge on the capacitor becomes Q = CV = 108pF × 100VQ = 108 ×...
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We know, capacitance = dielectric constant × C₀C = KC₀ = 6 × 18pF = 108pF [ see question - 2.8](a) if the voltage supply remained connected , then the potential difference across the capacitor will remain the same. e.g., V = 100V and hence, charge on the capacitor becomes Q = CV = 108pF × 100VQ = 108 × 10⁻¹² × 100 = 1.08 × 10⁻⁸ C (b) if the voltage supply was disconnected , then charge on the capacitor reamins the same. e.g., Q = 1.8 × 10⁻⁹C [ see question - 2.8]And hence the potential difference across the capacitor becomesV = Q/C = 1.8 × 10⁻⁹/1.08 × 10⁻¹⁰ = 16.6V read less
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Famous UNACADEMY PHYSICS TEACHER (use code ROHIT10 for 10% discount on unacademy)

We know that Capacitance will change if we insert any dielectric material between the plates. It becomes 'K' times the initial value, where K is the dielectric constant.Just remember the famous equation Q=CV, nowcase-1 when voltage supply remains connected, the potential difference across the capacitor...
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We know that Capacitance will change if we insert any dielectric material between the plates. It becomes 'K' times the initial value, where K is the dielectric constant.Just remember the famous equation Q=CV, nowcase-1 when voltage supply remains connected, the potential difference across the capacitor remains unchanged, therefore charge on capacitor has to change and becomes (k times) the initial charge because of capacitance 'C' changes to 'KC'.Case-2 When the capacitor is disconnected from voltage supply, then the circuit will become open so no charge can flow from one plate to another, hence charge re, mains same in this case and potential difference changes and reduces by a factor of 'K' (Can be easily deduced using the famous equation of Q=CV) read less
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Tutor in med/engg coaching centre for years

a)there you can observe a momentarily charge flow. While u insert a dielectric the capacitance get increased that times as the dielectric constant . As Q=CV when C increases and V remains the same charge stored in it get increased b) after the supply get disconnected and then inserting the dielectric...
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a)there you can observe a momentarily charge flow. While u insert a dielectric the capacitance get increased that times as the dielectric constant . As Q=CV when C increases and V remains the same charge stored in it get increased b) after the supply get disconnected and then inserting the dielectric also increase the capacitance while the charge cannot be increased since the circuit is open there by it would reduce the potential difference across the capacitance plates read less
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(a) Dielectric constant of the mica sheet, k = 6 Initial capacitance, C = 1.771 × 10 -11 F New Capacitance, C'= kC= 6x1.771x10-11 =106 pF Supply voltage, V = 100 V New Charge, q'=C'V=6x1.771x10-9 =1.06x 10-8 C Potential across the plates remains 100 V (b) Dielectric constant, k = 6 Initial...
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(a) Dielectric constant of the mica sheet,k= 6 Initial capacitance, C = 1.771 × 10-11F New Capacitance,C'=kC= 6x1.771x10-11=106 pF Supply voltage,V= 100V New Charge,q'=C'V=6x1.771x10-9=1.06x 10-8C Potential across the plates remains 100 V (b) Dielectric constant,k= 6 Initial capacitance, C = 1.771 × 10-11F New Capacitance,C'=kC= 6x1.771x10-11=106 pF If supply voltage is removed, then there will be no effect on the amount of charge in the plates. Charge = 1.771 × 10- 9C Potential across the plates is given by, read less
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