Question

Please help with number 1. I think i need to solve for delta V first but I am not sure if that is right.

Physics 2B Fall 2016, Van der Noordaa: Homework 5 Due 9/27 Capacitors 1. Two parallel plates, each of area 7.00 cm2, are separated by 3.50 mm of water. The electric field between the plates has a magnitude of 2290N/C. (a) What is the magnitude of the potential difference between the plates? (b) What is the charge stored on each plate? If the voltage across the plates remains the same when the water is replaced by air what is the new charge stored on each plate? Combinations of Capacitors 2.Four capacitors, C,=15 .00uF, C, = 3.00uF, C)-4.00uF, and C, = 4.00ur , are connected together. (a) hat is the equivalent capacitance if they are hooked in parallel? (b) What is the equivalent capacitance if ey are hooked in series?

Answer 1

we know that

potential difference V = E.d

V = 2290*3.5*10^(-3)

V = 6.87 V

(b)

capacitance C = e0*A / d

C = (8.85*10^(-12)*7*10^(-4)) / 3.5*10^(-3)

C = 17.7*10^(-13) F

charge Q = C * V

Q = 17.7*10^(-13) * 6.87

Q = 1.21*10^(-11) C

(c)

if water is replaced by air

C' = C / k

where k = dielectric constant of water = 80.4

C' = 17.7*10^(-13) / 80.4 = 22*10^(-15) F

so Q' = 22*10^(-15) * 6.87

Q' = 1.51*10^(-13) C

answer