A 70 pF capacitor and a 280 pF capacitor are both charged to 3.6 kV. They are then disconnected from the voltage source and are connected together, positive plate to positive plate and negative plate to negative plate. Find the resulting potential difference across each capacitor.
kV (70 pF capacitor)
kV (280 pF capacitor)
Find the energy lost when the connections are made. J

0.636 kJ

Explanation:

The charge on any capacity, q = CV, thus,

The initial charge on the 70 pF capacitor is

q = Cv

q = 70*10^-12 * 3.6*10^3

q = 2.52*10^-7 C

The charge on the 280 pF capacitor is q = C*v

q = 280*10^-12 * 3.6*10^3

q = 1.008x10^-6 C

When they are connected as stated, the net total charge remaining will be 1.008*10^-6 - 2.52*10^-7 = 7.56*10^-7 C

Since the capacitors are in parallel, the equivalent capacitance will be 70 + 280 pF = 350 pF

Remember, q = CV, then V = q/C

V = 7.56*10^-7 C / 350*10^12 F

V = 2160 V

b) The energy before is 1/2 C*v²

E = 1/2 * 70*10^-12 * 3600² + 1/2 * 280*10^-12 * 3600²

E = 4.536*10^-4 J + 1.814*10^-3 J

E = 2.268 kJ

The energy After is 1/2 Cv²

E = 1/2 * 70*10^-12 * 2160² + 1/2 * 280*10^-12 * 2160²

E = 3.266*10^-4 J + 1.306*10^-3 J

E = 1.632 kJ

so the loss is 2.268 - 1.632 = 0.636 kJ

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2