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?
E = 32*10^5 N/C
= (surface charge density)/epsilon0
Solve for the plate area (of each plate) using
A = (total charge)/(surface charge density)
that doesnt make sense...
To find the area of each plate in a parallel plate capacitor, we can use the formula for the electric field between the plates:
E = σ / ε₀,
where E is the magnitude of the electric field, σ is the surface charge density on each plate, and ε₀ is the permittivity of free space.
We can rearrange the formula to solve for σ:
σ = E * ε₀.
Given that the magnitude of the electric field E is 3.0 × 10^5 N/C and the magnitude of the charge Q is 0.17 µC, we can use the formula:
σ = Q / A,
where A is the area of each plate.
Now, we can substitute the values into the equation:
E * ε₀ = Q / A.
Rearranging the equation to express A in terms of Q, E, and ε₀:
A = Q / (E * ε₀).
The permittivity of free space, ε₀, is a constant with a value of approximately 8.85 × 10^(-12) C²/N·m².
Now, we can substitute the values and calculate the area of each plate:
A = (0.17 × 10^(-6) C) / (3.0 × 10^5 N/C * 8.85 × 10^(-12) C²/N·m²).
Simplifying the expression:
A = (0.17 × 10^(-6) C) / (3.0 × 10^5 N/C * 8.85 × 10^(-12) C²/N·m²).
A = (0.17 × 10^(-6) C) / (3.0 × 10^5 N/C * 8.85 × 10^(-12) C²/N·m²).
A ≈ (0.17 × 10^(-6) C) / (3.0 × 10^5 N/C * 8.85 × 10^(-12) C²/N·m²).
A ≈ 6.1 × 10^(-6) m².