There are 2 square parallel conducting plates separated by 0

There are 2 square parallel conducting plates separated by 0.5 cm with a length of 10cm on one side. The total charge on the positive plate is +10 muC. Determine the following: Electric field between the square plates Voltage, capacitance, and energy storage Voltage, capacitance, and energy storage after a dielectric with k = 2.8 is inserted between the two plates

Solution

here,

length of the side , a = 0.1 m

distance between the plates , d = 0.005 m

charge , Q = 10 uC

Q = 10^-5 C

capacitance , C = area *epsilon0 /d

C = 0.01 * 8.85 * 10^-12 /0.005

C = 1.77 * 10^-11 F

a.

Voltage , V = Q/C

V = 10^-5 /(1.77 * 10^-11)

V = 5.65 * 10^5 V

electric feild , E = V/d

E = 5.65 * 10^5 /0.005

E = 1.13 * 10^8 N/C

the electric feild between the square plates is 1.13 * 10^8 N/C

b.

energy stored , En = 0.5 * Q^2/C

En = 0.5 * (10^-5)^2 /(1.77 * 10^-11)

En = 2.82 J

the energy stored in the capacitor is 2.82 J

c.

when dielectric , k = 2.8

capacitance , C\' = area *k*epsilon0 /d

C\' = 0.01 * 8.85 * 10^-12 *2.8/0.005

C\' = 4.96 * 10^-11 F

Voltage , V\' = Q/C\'

V = 10^-5 /(4.96 * 10^-11)

V = 2.02 * 10^5 V

energy stored , En\' = 0.5 * Q^2/C\'

En\' = 0.5 * (10^-5)^2 /(4.96 * 10^-11)

En\' = 1.01 J

the energy stored in the capacitor is 1.01 J