A parallel-plate air capacitor has a capacitance of 810 pF. The charge on each p

Question

A parallel-plate air capacitor has a capacitance of 810 pF. The charge on each plate is 1.70 C. (a) What is the potential difference between the plates? 2098.77 V (b) If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? 4196 (c) How much work is required to double the separation? 1.7e-6 + 0/2 points | Previous Answers YF14 24.P.026 My Notes Ask Your Teacher A parallel-plate vacuum capacitor has 7.16 J of energy stored in it. The separation between the plates is 3.90 mm. If the separation is decreased to 1.50 mm, what is the energy stored if the following events occur? (a) the capacitor is disconnected from the potential source so the charge on the plates remains constant 3.58 (b) the capacitor remains connected to the potential source so the potential difference between the plates remains constant. 7.16

Explanation / Answer

here,

capcitance , C = 810 pF

charge , Q = 1.7 uC

V = Q/C = 2099 V

when the sepration is doubled

capacitance , C' = C/2

the work done , W = 0.5 * V^2 * (C - C' )

W = 0.5 * 2099^2 * 810/2

W = 8.92 * 10^-4 J

the work done required is 8.92 * 10^-4 J

--------------------------------------

E = 7.16 J

d = 3.9 mm , d' = 1.5 mm

a)

when the charge remains the same

the new energy , E' = 0.5 * Q^2 /C'

E' = 0.5 * Q^2 /( area * e0 /d')

E' = 0.5 * Q^2 /( area * e0 /1.5 )

E' = 0.5 * Q^2 /( area * e0 * (3.9/1.5) /3.9 )

E' = 0.5 * Q^2 /( area * e0 * (3.9/1.5) /d )

E' = 1.5 /3.9 * E

E' = 2.75 J

b)

when the potential remains the same

the new energy , E' = 0.5 * C' * V^2

E' = 0.5 * ( area * e0 /d') * V^2

E' = 0.5 * ( area * e0 * (3.9/1.5) /d ) * V^2

E' = 3.9 /1.5 * E

E' = 18.6 J

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