A parallel-plate air capacitor has a capacitance of 940 pF . The charge on each

ID: 2002800 • Letter: A

Question

A parallel-plate air capacitor has a capacitance of 940 pF . The charge on each plate is 2.5 C .

Part A

What is the potential difference between the plates?

Part B

If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled?

Part C

How much work is required to double the separation?

A parallel-plate air capacitor has a capacitance of 940 pF . The charge on each plate is 2.5 C .

Part A

What is the potential difference between the plates?

V = V

Part B

If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled?

V = V

Part C

How much work is required to double the separation?

W = mJ

we use the formula Q=CV

we have C=940pF, Q=2.5uC

we will find C

C=EA/d E is pemitivity,A is area of plate and d is distance between plates

so when d is doubled capacitance is halved henceand potential is doubled

energy can be calculated by (1/2)Cv^2

and we can find energy before seperating and after seperating

and then subtract them so we get -ve of work done in this seperation

solution:
1.

C = q / V

V = 2.5*10^-6 / (940*10^-12) = 2659.57 V

if the separation is doubled then

C = k*A/d => C halved => V(potential difference) doubled

V = 5319.14 V

U = q^2/2C => U2 = 2*U1 => U = U1 = 1/2*CV^2 = 1/2*(Q^2/C) = (2.5*10^-6)^2 / (2*940*10^-12) = 0.0033 J

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.