Please help me with the 2 problem sets listed here. I need the answers please an
ID: 1576167 • Letter: P
Question
Please help me with the 2 problem sets listed here. I need the answers please and completely.
A parallel plate, air filled capacitor, has plates of area 0.80 m2 and a separation of 0.050 mm. (a) Find the capacitance. (b) If a voltage of 25 volts is applied to the capacitor, what is the magnitude of the charge on each plate? (c) What is the energy stored in the capacitor? nF microc microJ Now the plates are filled with strontium titanate having a dielectric constant of 310 and a dielectric strength of 8.0 kV/mm. (d) What is the new value of capacitance? (e) For a voltage of 25 V, what is the charge on each plate? (f) What is the energy stored in the capacitor now? () What is the maximum voltage that can be placed across the capaor? microF microC micro]Explanation / Answer
Given
Parallel plate capacitor with
Area A = 0.80 m^2 , separation d = 0.050 m
we knwo that the capacitance of the parallel plate capacitor is
a) C = epsilon not*A/d
epsilon not is perimitivity of free space
C = (8.854*10^-12*0.80)/(0.05) F
C = 1.41664*10^-10 F
C = 0.141664 nF
b) from relation Q = C*V
Q = 1.41664*10^-10 *25 C
Q = 3.5416*10^-9 C
Q = 0.0035416*10^-6 C
Q = 0.0035416 micro C
c) energy stored is
U = 0.5*C*V^2 = 0.5*1.41664*10^-10*25^2 J = 4.427*10^-8 J
if dielectric material placed / filled the separation of the plates then
k = 310 and its strength is 8000 V/mm
d) new value of capacitance is c' = k*c = 310*1.41664*10^-10 F
C' = 4.391584*10^-8 F
C' = 0.04391584*10^-6 F
e) charge on each plate is Q = C'*V = 0.04391584*10^-6*25 C = 1.097896*10^-6 C= 1.1 microC
f) energy stored in the capacitor is
U = 0.5*C'*v^2
U = 0.5*0.04391584*10^-6*25^2 J
U = 13.7237*10^-5 J
g) maximum voltage is 8000 V/mm