The magnitude of the electric field between two plates of a parallel plate capacitor is 2.4×10^5 N/C. Each plate carries a charge whose magnitude is o.15 mC. What is the area of each plate?
E= Q/area * 1/epsilon
okay so I did that, but I still did not end up with the correct answer. The book says that the area of each plate should be 7.1×10^-2 m^2, but I keep getting an answer that is completely off from that.
area= Q/Ee= .15E-3/(2.4E5*8.85E-12)=71m^2
Is the mC milliCoulombs, or microCoulombs?
It microCoulombs, then area= .071m^2, which agrees with your text.
Thank you so much. I've got an AP test tomorrow and that really helps :)
To find the area of each plate, we can use the formula for the electric field between the plates of a parallel plate capacitor:
E = σ / ε₀
where E is the magnitude of the electric field, σ is the surface charge density, and ε₀ is the permittivity of free space.
In this case, we are given the magnitude of the electric field, E = 2.4×10^5 N/C. We are also given the charge on each plate, which can be written as q = 0.15 mC = 0.15 × 10^(-3) C.
The surface charge density can be calculated as σ = q / A, where A is the area of each plate.
Substituting these values into the formula for the electric field, we have:
2.4×10^5 N/C = (0.15 × 10^(-3) C) / A
Solving for A, we find:
A = (0.15 × 10^(-3) C) / (2.4×10^5 N/C)
A = 6.25×10^(-10) m^2
Therefore, each plate has an area of 6.25×10^(-10) square meters.