The magnitude of the electric field between the plates of a parallel plate capacitor is 3.0X10^5 N/C. Each plate carries a charge whose magnitude is 0.17 µC. What is the area of each plate?
can someone please work with it in numbers!!
Isn't E= chargedenstiy/epsilion
and charge denstiy= charge/area ?
To find the area of each plate, you can use the formula for the electric field between the plates of a parallel plate capacitor, which is given by:
E = V/d
Where:
E is the electric field magnitude (given as 3.0 × 10^5 N/C),
V is the voltage across the capacitor plates,
d is the separation between the plates.
First, let's convert the charge magnitude to coulombs:
0.17 µC = 0.17 × 10^-6 C
Next, we need to calculate the voltage V using the formula:
V = Q / C
Where:
Q is the charge on each plate (0.17 × 10^-6 C),
C is the capacitance of the capacitor.
To find the capacitance, we can use the formula:
C = ε₀(A/d)
Where:
ε₀ is the permittivity of free space (approximately 8.85 × 10^-12 F/m),
A is the area of each plate.
Rearranging the formula for capacitance, we get:
A = Cd / ε₀
Now, let's substitute the given values into the formulas and solve step by step:
1. Calculate V:
V = (0.17 × 10^-6 C) / C
V = 0.17 × 10^-6 V
2. Calculate the capacitance C:
C = ε₀(A/d)
C = (8.85 × 10^-12 F/m) * (d / 1 m)
C = 8.85 × 10^-12 d F
3. Calculate the area A:
A = Cd / ε₀
A = (8.85 × 10^-12 d F) * (d / 1 m) / (8.85 × 10^-12 F/m)
A = d^2 m²
By substituting the known value for the electric field (E = 3.0 × 10^5 N/C), we can solve for the separation between the plates (d):
E = V/d
3.0 × 10^5 N/C = (0.17 × 10^-6 V) / d
d = (0.17 × 10^-6 V) / (3.0 × 10^5 N/C)
Now, substitute the calculated value of d back into the equation for the area:
A = d^2 m²
A = [(0.17 × 10^-6 V) / (3.0 × 10^5 N/C)]^2 m²
After performing the calculations, you should find the area (A) of each plate in square meters.