A capacitor is contructed out of 21 coines facing each other in air at a distanc

Question

A capacitor is contructed out of 21 coines facing each other in air at a distance of 0.5 mm and connected to a voltage source of Vs=9V.

1. If the medium between the coins is air and its capacitance is 7.52pF, what is the energy stored in the capacitro and what is the diameter of a coin? (Permittivity of Vacuum: e0=8.854x10-12 F/m)

2. If the capacitor is immersed in disitilled water, what is the capacitance and the energy stored in the capacitor? (Relative Permittivity of distilled water: er=80)

3. In the case of air, if the coins are replaced by difference coins, the capacitance becomes 9.22pF. Are these new coins larger or smaller than the previous coins. Explain.

Please help!

Explanation / Answer

n = number of capator formed by 21 coins = 20

C = capacitance of single capacitor = 7.52 pF

Hence   , 20 capacitors are connected in series ,

Ceq = equivalnet capacitance of 20 capacitors connected in series

         = C/20

        = 7.52/20

        = 0.376 pF

1) E = energy stored in capacitor = Ceq * Vs^2 /2 = 0.376 * 9^2/2 = 15.22*10^12 j

2) if cacitor in imerser in water , then

     ceq1 = equivalent capacitance = er* 0.376 = 30.08 pF

     E1 = energy stored = Ceq1 * Vs^2 /2 = 30.08 * 9^2/2 =1218 *10^-12 J

3) C = capacitance = eo* A/d

      where A = surface area .

                d = separation between plates ,

so   if C increases , then A will increase

As , new capatitance is large , So A is increased,

hence coins are larger than previous one ,

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