A capacitor is constructed with two parallel metal plates each with an area of 0.61 m^2 and separated by d = 0.80 cm. The two plates are connected to a 5.0-volt battery. The current continues until a charge of magnitude Q accumulates on each of the oppositely charged plates. Find the charge Q.
I'm stuck, please help
the second question the above question is Find the capacitance of the parallel plates.
I have to use this formula to find Q Q = CΔV, and C = (ε0*A)/d
These are what I did but don't know what ε0 is
C= (ε0*0.61 m^2 )/.0080m= stuck here. Is there anybody know what to do from here?
ε0=8.85*10^-12
C=[(8.85*10^-12)*.61]/.008=
Q=([(8.85*10^-12)*.61]/.008=)*5=
Parallel Plate Capacitor Capacitance Calculator for second question
www.daycounter.com/Calculators/Plate-Capacitor-Calculator.phtml
To find the charge Q accumulated on each plate of the capacitor, we need to use the formula for the capacitance of a parallel-plate capacitor.
The capacitance of a parallel-plate capacitor is given by the equation:
C = ε₀ * A / d
Where:
C is the capacitance of the capacitor,
ε₀ is the permittivity of free space (a constant value),
A is the area of one of the plates, and
d is the distance between the plates.
To find the charge Q, we can rearrange the equation to solve for Q:
Q = C * V
Where:
Q is the charge,
C is the capacitance, and
V is the voltage applied across the capacitor.
Let's substitute the values given into the equations to find the charge Q:
Given:
Area of each plate (A) = 0.61 m^2
Distance between plates (d) = 0.80 cm = 0.0080 m
Voltage (V) = 5.0 volts
Let's calculate the capacitance (C):
C = ε₀ * A / d
The permittivity of free space (ε₀) is a constant value equal to 8.85 x 10^-12 F/m.
C = (8.85 x 10^-12 F/m) * (0.61 m^2) / (0.0080 m)
C ≈ 6.7598 x 10^-9 F
Now, let's calculate the charge (Q):
Q = C * V
Q = (6.7598 x 10^-9 F) * (5.0 volts)
Q ≈ 3.3799 x 10^-8 Coulombs
Therefore, the charge accumulated on each plate (Q) is approximately 3.38 x 10^-8 Coulombs.