a) C = q/V so q = C V
= 5*10^-6 * 1.5
= 7.5 * 10^-6 Coulombs
now do E = (1/2) C (1.5)^2 to get energy
E = .5 (5*10^-6)(2.25)
E = 5.63*10^-6 Joules
b) Luckily we derived this in your last question, but now that I see your questions you probably have it in your book and have not had calculus so
E = (1/2) C V^2 from your book or the other problem
E = (1/2) C (q/C)^2
(q/C)^2 = 2 E/C
q^2/C^2 = 2 E/C
q^2 = 2 E C
q = sqrt (2E*C)
q = sqrt (2 * 5 * 10*-6)
q = 10^-3 sqrt (10)
q = 3.16 * 10^-3 coulombs
c)V = q/C = 2.24*10^-3 / 5*10^-6
V = .447* 10^3 = 447 volts, watch out
(a) How much charge does a battery have to supply to a 5.0 uF capacitor to create a potential difference of 1.5 V across its plates? How much energy is stored in the capacitor in this case? (b) How much charge would the batter have to supply to store 1.0 J of energy in the capacitor? What would be the potential across the capacitor in this case.
Once again, I am lost!!!
2 answers
Good but quite difficult to understand.