Can anyone help me with a physics capacitance and potential difference problem?

Question

I believe that a)0 b)0 c)20V, but please check my answers if you know how to do this. Any help is greatly appreciated.

Two metal objects have net charges of +70 pC and -70pc, which result in a 20V potential difference between them.
(a)What is the capacitance of the system?
(b)If the charges are changed to +200pc and -200pc, what does the capacitance become?
(c)What does the potential difference become?

Answer 1

The equation that you need is Q = C V
or in words: charge = capacitance * voltage

The answer to part (a) is: 70 pC / 20 V = 3.5 pF

For part (b), the question is not very clear. Do they want us to change the amount of charge to 200 pC, and also change the capacitance just the right amount to keep the potential at 20 V?
If that's really what they are asking, then the answer is:
200 pC / 20 V = 10 pF

Part (c) is unclear also. Are they saying to keep the capacitance the same as in part (a), but change the charge to +- 200 pC, like in part (b)?
If that's the question, then the answer is:
200 pC / 3.5 pF = 57.1 V

Answer 2

we use the formulation Q=CV ..we've C=920pF Q=2.fifty 5 we can discover C C=EA/d E is pemitivity,A is component of plate and d is distance between plates so whilst d is doubled capacitance is halved henceand potential is doubled capacity could nicely be calculated by potential of (a million/2)Cv^2 and we are in a position to discover capacity till now seperating and after seperating and then subtract them so we get -ve of artwork finished in this seperation..... answer: a million. C = q / V => V = 2.fifty 5*10^-6 / 920*10^-12 = 2771.seventy 4 V 2. C = ok*A/d => C halved => V(potential distinction) doubled 3. U = q^2/2C => U2 = 2*U1 => ?U = U1 = (2.fifty 5*10^-6)^2 / 2*920*10^-12 = 0.0035 J