A 40-pF capacitor is charged to a potential difference of 500 V. Its terminals are then connected to those of an uncharged 10-pF capacitor. Calculate: (a) the original charge on the 40-pF capacitor; (b) the charge on each capacitor after the connection is made; and (c) the potential difference across the plates of each capacitor after the connection.

a)  

Q = CV

Q = (40pF) * (500V) = 20000 pC = 20 nC

b)

Cp = C1 + C2

Cp = 40 pF + 10pF = 50 pF

Q = CV

V = Q/C = 20000 pF/50pF = 400V

Q1 = C1V1

Q1 = 40 pF * 400 = 16000pC = 16 nC

Q2 = C2V2 = 10 pF * 400 = 4000 pC = 4 nC

c) 400V

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

answer///scientific method;