Question
The following 9 questions refer to a parallel plate capacitor, Initially with air as the dielectric. The plates have an area of 1.0 m^2 and they are separated by 4 cm. One plate is connected to the +400 V terminal of a power supply while the other is connected to the grounded (zero V) terminal. 5. What is the capacitance of the capacitor? 6. What charge (magnitude and sign) is on the plate connected to the +400 V terminal. 7. What charge (magnitude and sign) is on the plate connected to the 0 V terminal? 8. What is the value of the electric field at a point half-way between the plates? 9. What is the value of the electric potential at a point half-way between the plates? 10. What is the value of the electric field at a point which lies 1.0 cm from the + plate and 3.0 cm from the other plate? 11. How much work is done on a charge of -9.62 times 10^-9 C when it is moved a distance of 4.00 cm Parallel to the + plate? 12. Suppose the capacitor is now immersed in glycerol (K = 45.0). What charge (magnitude and sign) is now on the plate connected to the +400 V terminal? 13. If you charge the air-dielectric capacitor to 400 V, disconnect it from power supply, and then immerse it in glycerol, what voltage will you observe between its plates?
Explanation / Answer
given
A = 1 m^2
d = 4 cm = 0.04 m
5) C = A*epsilon/d
= 1*8.854*10^-12/0.04
= 2.2*10^-10 F
6) Q = C*V
= 2.2*10^-10*400
= +8.8*10^-8 C
7) Q = -8.8*10^-8 C
8) E = V/d
= 400/0.04
= 1*10^4 N/c
9) at a point half-way = 400/2
= 200 Volts
10) E = 1*10^4 N/c
11) zero.
12) Q' = k*Q
= 45*8.8*10^-8
= +3.96*10^-6 C
13) V' = V/k
= 400/45
= 8.89 volts