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?
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.
To solve these questions, we will use the principles of Coulomb's law and the equations for electric field, potential energy, voltage, and capacitance.
a) The electric field (E) between the plates of a parallel plate capacitor is given by E = σ/ε0, where σ is the surface charge density and ε0 is the permittivity of vacuum.
First, we need to find the surface charge density. The force (F) experienced by the test charge is given by F = qE, where q is the test charge. Rearranging the equation, we get E = F/q.
Substitute the given values: F = 0.0320 N and q = +3.00 mC (converted to Coulombs: 3.00e-3 C).
E = 0.0320 N / (3.00e-3 C)
E ≈ 10.67 N/C
Therefore, the value of the electric field in the capacitor is approximately 10.67 N/C.
b) The change in potential energy (ΔU) of a test charge moving between the plates of a capacitor is given by ΔU = qΔV, where q is the test charge and ΔV is the change in voltage.
Substitute the given values: q = +3.00 mC (converted to Coulombs: 3.00e-3 C).
ΔU = (3.00e-3 C) ΔV
Since we do not have the value of ΔV, we cannot determine the change in potential energy without further information.
c) The change in voltage (ΔV) of the capacitor is given by ΔV = Ed, where E is the electric field and d is the distance between the plates.
Substitute the given values: E = 10.67 N/C and d = 0.0400 m.
ΔV = (10.67 N/C) (0.0400 m)
ΔV = 0.4268 V
Therefore, the test charge moved through a change in voltage of 0.4268 V.
d) The voltage (V) of the battery that charged the capacitor is equal to the change in voltage.
V = ΔV
V = 0.4268 V
Therefore, the voltage of the battery that charged the capacitor is 0.4268 V.
e) The capacitance (C) of a capacitor is given by C = ε0(A/d), where ε0 is the permittivity of vacuum, A is the area of the plates, and d is the distance between the plates.
Substitute the given values: ε0 = 8.85e-12 C²/Nm², A = 0.250 m², and d = 0.0400 m.
C = (8.85e-12 C²/Nm²)(0.250 m² / 0.0400 m)
C = 5.53125e-11 C²/N
To convert the capacitance to picofarads, divide by 1e-12:
C ≈ 55.3125 pF
Therefore, the capacitance of the capacitor is approximately 55.3125 picofarads.