Answer:
Explanation:
Capacitance of the capacitor = 13.5μF
Voltage across plate is 24V
Dielectric constant k=3.55.
a. Energy in capacitor is given by
E=1/2CV^2
We want to calculate energy without the dielectric substance
Given that C=13.5 μF and V=24V
The capacitance give is with dielectric so we need to remove it
C=kCo
Co=C/k
Then the Co=13.5μF/3.55
Co=3.803μF
Then
E=(1/2)×3.803×10^-6×24^2
E=1.1×10^-3J
E=1.1mJ
b. Energy in capacitor is given by
E=1/2CV^2
The capacitance given is with a dielectric, so we are going to apply it direct.
Given that C=13.5 μF and V=24V
Then
E=(1/2)×13.5×10^-6×24^2
E=3.89×10^-3J
E=3.9mJ
c. The energy without dielectric is 1.1mJ and the energy with dielectric is 3.9mJ
The energy increase when the dielectric material is added
d. Dielectrics in capacitors serve three purposes: to keep the conducting plates from coming in contact, allowing for smaller plate separations and therefore higher capacitances;
Therefore, Since dielectric allow higher capacitance, and energy of a capacitor is directly proportional to the capacitance, then the higher the capacitance the higher the energy.
