I thought this problem was going to be a cakewalk, now I am far too confused to
ID: 1433927 • Letter: I
Question
I thought this problem was going to be a cakewalk, now I am far too confused to know what to do. It seems like a simple capacitor problem gone wrong- I tried the basic method of solving it, and then involving a dialectic K=1.00059 (which doesn't change much). For problem a) I had the result C = 2.3e-11 and this answer and all subsequent answers are already considered wrong by the computer. Any clarification would be appreciated.
A parallel-plate air capacitor is made by using two plates 14 cm square, spaced 5.3 mm apart. It is connected to a 6-V battery ) What is the capacitance? (b) What is the charge on each plate? c) What is the electric field between the plates? What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of 10.6 mm, what are the answers to parts (a)-(d)? capacitance charge electric field energyExplanation / Answer
here,
side of square , a = 0.14 m
spaced , d = 5.3 * 10^-3 m
voltage , V = 6 V
(a)
capacitance , C= A * e0 /d
C= pi * (0.14)^2 * 8.85 * 10^-12 /( 5.3 * 10^-3)
C = 1.03 * 10^-10 F
(b)
the charge on each plate , Q = C * V
Q = 1.03 * 10^-10 * 6
Q = 6.17 * 10^-10 C
(c)
the electric feild between the plates , E = V /d
E = V /( 5.3 * 10^-3)
E = 1132.1 N/C
d)
the energy stored in the capacitor, E = 0.5 * C * V^2
E = 0.5 * 1.03 * 10^-10 * 6^2
E = 1.85 * 10^-9 J