Two 50 F capacitors connected in parallel are charged by using a 12.0 V battery. As soon as the battery is removed, then an insulator of dielectic constant = 5.0 is inserted to the filling spaces between these capacitors.
Calculate the energy stored in the capacitor system after insertion of insulators between two capacitors
Answers
Thus the energy stored in the capacitor system after insertion of insulators between two capacitors is 7.2 mJ
Step-by-step explanation:
Using formula of parallel :
C = C1 + C2
Put the value into the formula
C = 50 + 50 = 100 mu F
We need to calculate the charge
Using formula of potential difference
V = Q / C
Q = V C
Put the value into the formula
Q = 12.0 x 100 x 10^-6
Q = 12 x 10^-4 C
Now using formula of energy :
U= 1 /2 x Q^2 / C
Put the value into the formula
U = 1 / 2 x 1 / (2 x 10^-4)^2 / 100 x 10^-6
U = 0.0072 J
U = 7.2 mJ
Thus the energy stored in the capacitor system after insertion of insulators between two capacitors is 7.2 mJ
Answer:
Thus the energy stored in the capacitor system after insertion of insulators between two capacitors is 7.2 mJ
Step-by-step explanation:
Using formula of parallel
:
C = C1 + C2
Put the value into the formula
C = 50 + 50 = 100 mu F
We need to calculate the charge
Using formula of potential difference
V = Q / C
Q = V C
Put the value into the formula
Q = 12.0 x 100 x 10^-6
Q = 12 x 10^-4 C
Now using formula of energy
:
U= 1 /2 x Q^2 / C
Put the value into the formula
U = 1 / 2 x 1 / (2 x 10^-4)^2 / 100 x 10^-6
U = 0.0072 J
U = 7.2 mJ
Thus the energy stored in the capacitor system after insertion of insulators between two capacitors is 7.2 mJ
Step-by-step explanation:
Hope this answer will help you.✌️