Question

A q = +3.00 mC test charge is placed inside a constant uniform electric field created in a parallel plate capacitor connected to a battery of unknown voltage.
 
The capacitor is made of two parallel plates each with area equal to 0.250 m2 separated by a distance of 0.0400 m and with no dielectric other than vacuum. The permitivitty of vacuum is 8.85e-12 C2/Nm2.

a) If it takes 0.0320 N of force to hold the charge the stationary between the plates, what is the value of the electric field in the capacitor?
b) What is the change of potential energy of the test charge moves without assistance from one plate to the other? (note: "+" would indicated a GAIN in potential energy)
c) Through what change in voltage did the test charge move?
d) What is the voltage of the battery that charged the capacitor?
e) What is the capacitance of the capacitor in picofarads?

Answer 1

E=F/q=0.032/0.003=10.67 V/m,

E= - grad φ
E=Δφ/Δx= Δφ/d =>
Δφ=Ed= 10.67•0.04=0.427 V.

ΔPE = - Work of electric field=
=-q•Δφ=-0.003•0.427= -1.28•10⁻³ J.
Δφ=V.
C=εε₀A/d,
ε=1,
ε₀=8.85 •10⁻¹² F/m,
C=8.85•10⁻¹²•0.25/0.04=5.53•10⁻¹¹F=
=55.3 pF.