The plates of an air-filled parallel-plate capacitor with a plate area of 15.5 cm2 and a separation of 9.15 mm are charged to a 150-V potential difference. After the plates are disconnected from the source, a porcelain dielectric with κ = 6.5 is inserted between the plates of the capacitor.(a) What is the charge on the capacitor before and after the dielectric is inserted?Qi= CQf= C(b) What is the capacitance of the capacitor after the dielectric is inserted?F(c) What is the potential difference between the plates of the capacitor after the dielectric is inserted?V(d) What is the magnitude of the change in the energy stored in the capacitor after the dielectric is inserted?J

(a)Qi = Qf = 2.16×10-¹⁰C

(b) C = 9.36×10-¹² F

(c) V = 23.1V

(d) ΔU = -1.37×10‐⁸ J

Explanation:

Please see the attachment below.

The chargees on the capacitor plates is the same in both cases of inserting and not inserting the dielectric. The presence of a dielectric reduces the voltage across the plates but increases the capacitance of the capacitor.

See full solution below.

the answer would be c

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.