Answer:
- E = 1.25*10^3 N/C
- Q = 1.08*10^-12 C
- Q = 8.69*10^-11 C
Explanation:
- In order to calculate the magnitude of the electric field between the plates, you use the following formula:
[tex]E=\frac{V}{d}[/tex] (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):
[tex]E=\frac{7.55V}{6.00*10^{-3}m}=1.25*10^3\frac{N}{C}[/tex]
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:
[tex]Q=CV[/tex] (2)
C: capacitance of the capacitor
The capacitance of a parallel plate capacitor is:
[tex]C=\epsilon_o \frac{A}{d}[/tex] (3)
You replace the previous formula into the equation (2) and solve for Q:
[tex]Q=(\epsilon_o k\frac{A}{d})(V)=(8.85*10^{-12}C^2/Nm^2)\frac{7.37*10^{-4}m^2}{6.00*10^{-3}m}\\\\Q=1.08*10^{-12}C[/tex]
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:
[tex]C=\epsilon_o k\frac{A}{d}[/tex] (4)
You replace the previous formula in the equation (2) and replace the values of all parameters:
[tex]Q=(\epsilon_o k\frac{A}{d})(V)=(8.85*10^{-12}C^2/Nm^2)(80.0)\frac{7.37*10^{-4}m^2}{6.00*10^{-3}m}\\\\Q=8.69*10^{-11}C[/tex]
The charges on each plate is 8.69*10^-11 C
