Answer:
- E = 1.25*10^3 N/C
- Q = 1.08*10^-12 C
- Q = 8.69*10^-11 C
Step-by-step explanation:
- In order to calculate the magnitude of the electric field between the plates, you use the following formula:
(1)
V: potential difference between the plates = 7.55V
d: distance between the plates = 6.00mm = 6.00*10^-3m
You replace the values of the parameters n the equation (1):
The magnitude of the electric field between the plates is 1.25*10^3N/C
- The charge on each plate is given by the following formula:
(2)
C: capacitance of the capacitor
The capacitance of a parallel plate capacitor is:
(3)
You replace the previous formula into the equation (2) and solve for Q:

The charge on each plate is 1.08*10^-12C = 1.08pC
- If water is placed in between the plates, the dielectric permittivity is changes by a factor of k = 80.0.
The capacitance of a parallel plate capacitor with a substance with a constant dielectric k, is given by:
(4)
You replace the previous formula in the equation (2) and replace the values of all parameters:

The charges on each plate is 8.69*10^-11 C