A3.00 μf capacitor is charged to 480 v and a 4.00 μf capacitor is charged to 500 v . part a these capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. what will be the potential difference across each capacitor?

Q before connected = Q after connected C1V1+C2V2 = (C1+C2) V

C1= 3×10^-6 F

V1= 480v

C2= 4×10^-6 F

V2= 500v

(3×10^-6)×(480) + (4×10^-6)×(500) = (3×10^-6 + 4×10^-6) × V

Simplifying the above, we get:

( 1440× 10^-6) + (2000 ×10^-6) = (7 × 10^-6) × V.

Further simplified as:

3440 × 10^-6 = 7 × 10^-6 × V

Making V the subject

V = 491.43volts

Therefore the potential difference across each capacitor is 491.43v

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.

a is the best answer your welcome