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Homework Statement:

The area of each plate of a parallel plate capacitor is 0.024 m2. The plates are 2.75 mm apart with a dielectric material (κ = 3.0) between them. The maximum possible electric field between the plates is 3.25e5 V/m.
(a) What is the charge on the plates?
Homework Equations:

C(dielectric) = Q/V(dielectric) , where V(dielectric) = 1/κ * V(original)
C(dielectric) = κ * C(original)
C = ε0 * (a/d) for parallel plate capacitor
This is an online HW question so maybe my digits are just off from rounding or something, but I don't know why I am not finding the correct answer. I got Q = 6.9e8 as the magnitude of charge on each plate.
I basically just needed to calculate the original capacitance of this capacitor using c = ε0*(a/d) , then multiply this by the dielectric constant ( κ = 3.0) since the capacitance of a capacitor with a dielectric is just the original capacitance without the dielectric, multiplied by the dielectric constant (so capacitance can be increased by a factor of (κ) when a dielectric is inserted between its plates).
Area = .024 m^2
Distance between plates = 2.75mm
C(dielectric) = ε0 (.024/2.75) *3 = 2.31e13
Now I need to find the voltage using V(dielectric) = (1/κ ) * V(original). We have a parallel plate capacitor which means our E field is uniform, therefore:
E = 3.25e5
V = E*D = 3.25e5 (2.75) = 893750
so, V(dielectric) = 893750(1/3) = 297916.6667
finally, we can use Q = C(dielectric) * V(dielectric) to get:
Q = 2.31e13 (297916.6667) = 6.9e8
I have no clue where I'm going wrong. I tried entering 7.0e8 and 6.8e8 incase it was a rounding issue and those were both incorrect as well. So either I'm just not working this problem correctly, or Webassign just sucks.
I basically just needed to calculate the original capacitance of this capacitor using c = ε0*(a/d) , then multiply this by the dielectric constant ( κ = 3.0) since the capacitance of a capacitor with a dielectric is just the original capacitance without the dielectric, multiplied by the dielectric constant (so capacitance can be increased by a factor of (κ) when a dielectric is inserted between its plates).
Area = .024 m^2
Distance between plates = 2.75mm
C(dielectric) = ε0 (.024/2.75) *3 = 2.31e13
Now I need to find the voltage using V(dielectric) = (1/κ ) * V(original). We have a parallel plate capacitor which means our E field is uniform, therefore:
E = 3.25e5
V = E*D = 3.25e5 (2.75) = 893750
so, V(dielectric) = 893750(1/3) = 297916.6667
finally, we can use Q = C(dielectric) * V(dielectric) to get:
Q = 2.31e13 (297916.6667) = 6.9e8
I have no clue where I'm going wrong. I tried entering 7.0e8 and 6.8e8 incase it was a rounding issue and those were both incorrect as well. So either I'm just not working this problem correctly, or Webassign just sucks.