A parallel plate capacitor has square plates with sides of length 10 cm. The dis
ID: 2236631 • Letter: A
Question
A parallel plate capacitor has square plates with sides of length 10 cm. The distance between the plates is 4 mm. The plates are charged up to 35 volts. 1. What is the electric field between the plates? 2. What is the amount of charge on each plate? 3. What is the capacitance? 4. What is the energy stored by the capacitor? 5. The space between the plates is filled with a dielectric which has a dielectric constant of 20, and the new capacitor is charged up to the same voltage as before. How much energy is stored in the capacitor now? Any help would be great!Explanation / Answer
A = 100* 10^ -4 m^2 = 10^ -2 m^2
d = 4mm = 4* 10^ -3 m
capacitance = A *epsilon0 /d = 10^ -2 * 8.85 * 10^ -12/ (4* 10^ -3)
=> C = 2.2125 * 10^ -11 Farad
charge stored Q = CV = 2.2125 * 10^ -11 *35 = 7.74375 * 10^ -10 coulomb
energy = 0.5 C*V^2 = 1.35 * 10^-8 joules
if the dielectric constant of the medium = 20,
then capacitance will be 20 times the previous value and so will be the energy stored.
so, E = 27 * 10^ -8 j
electric field = V/d = 35/ 4* 10^-3 = 8750 V/m