A parallel plate air capacitor of capacitance 245pF has a charge of magnitude 0.148ìC on each plate. The plates are 0.328mm apart.
(a) What is the potential difference between the plates?
(b) What is the area of each plate?
(c) What is the electric field magnitude between the plates?
(d) What is the surface charge density on each plate?
You are so dumb. This is supposed to be for people who need help.
uh, no. They're not "dumb". You're just mad that they called you out on cheating,which you like to call "asking for help"(help that you could just get from asking your teacher).
To find the answers to the given questions, we can use the formulas and equations related to capacitance and the properties of parallel plate capacitors. Let's go through each question one by one.
(a) To find the potential difference between the plates, we can use the formula:
V = Q / C
where V is the potential difference, Q is the charge, and C is the capacitance.
Given:
Charge on each plate (Q) = 0.148 μC
Capacitance (C) = 245 pF
First, let's convert the charge from microcoulombs to coulombs:
0.148 μC = 0.148 x 10^(-6) C
Now, substitute the values into the formula:
V = (0.148 x 10^(-6) C) / (245 x 10^(-12) F)
= 0.148 / 245 x 10^(-6+12) V
= 0.148 / 245 x 10^(-6-12) V
= 0.148 / 245 x 10^(-18) V
≈ 0.603 V (rounded to three decimal places)
Therefore, the potential difference between the plates is approximately 0.603 V.
(b) To find the area of each plate, we can use the formula:
C = ε₀ * (A / d)
where C is the capacitance, ε₀ is the vacuum permittivity, A is the area, and d is the distance between the plates.
Given:
Capacitance (C) = 245 pF
Distance between the plates (d) = 0.328 mm
First, let's convert the capacitance from picofarads to farads:
245 pF = 245 x 10^(-12) F
Now, re-arrange the formula to solve for A:
A = C * d / ε₀
We need to know the value of ε₀, which is the vacuum permittivity. The value of ε₀ is approximately 8.85 x 10^(-12) F/m.
Now substitute the values into the formula:
A = (245 x 10^(-12) F) * (0.328 x 10^(-3) m) / (8.85 x 10^(-12) F/m)
= 245 x 10^(-12-12+(-3+12)) m^2 / 8.85 x 10^(-12)
= 245 x 10^(-3) m^2 / 8.85
≈ 27.683 m^2 (rounded to three decimal places)
Therefore, the area of each plate is approximately 27.683 m^2.
(c) To find the electric field magnitude between the plates, we can use the formula:
E = V / d
where E is the electric field magnitude, V is the potential difference, and d is the distance between the plates.
Given:
Potential difference (V) = 0.603 V
Distance between the plates (d) = 0.328 mm
First, let's convert the distance from millimeters to meters:
0.328 mm = 0.328 x 10^(-3) m
Now substitute the values into the formula:
E = (0.603 V) / (0.328 x 10^(-3) m)
= 0.603 / 0.328 x 10^(0+3) V/m
≈ 1840.854 V/m (rounded to three decimal places)
Therefore, the electric field magnitude between the plates is approximately 1840.854 V/m.
(d) To find the surface charge density on each plate, we can use the formula:
σ = Q / A
where σ is the surface charge density, Q is the charge, and A is the area.
Given:
Charge on each plate (Q) = 0.148 μC
Area of each plate (A) = 27.683 m^2
First, let's convert the charge from microcoulombs to coulombs:
0.148 μC = 0.148 x 10^(-6) C
Now substitute the values into the formula:
σ = (0.148 x 10^(-6) C) / (27.683 m^2)
= 0.148 / 27.683 x 10^(-6-2) C/m^2
≈ 5.35 x 10^(-9) C/m^2 (rounded to three decimal places)
Therefore, the surface charge density on each plate is approximately 5.35 x 10^(-9) C/m^2.