The plates of a parallel plate capacitor each have an area of 0.40 m 2 and are separated by a distance of 0.02 m. They are charged until the potential difference between the plates is 3000 V. The charged capacitor is then isolated.
-Suppose that a dielectric sheet is inserted to completely fill the space between the plates and the potential difference between the plates drops to 1000 V. What is the capacitance of the system after the dielectric is inserted?

Question

Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor with empty space between the plates is given by:
C = ε0A/d
where ε0 is the permittivity of free space, A is the area of each plate, and d is the distance between the plates. Plugging in the given values,
C = (8.85 × 10−12 F/m)(0.40 m2)/(0.02 m) = 1.77 × 10−9 F
When the dielectric is inserted, the capacitance increases by a factor of κ, the dielectric constant of the material:
C' = κC
The potential difference across the plates is inversely proportional to the capacitance, so we can use the equation for capacitors in series to find the new potential difference:
1/C' = 1/(κC) + 1/C
V' = V(1 - κ)/κ
where V is the original potential difference, and V' is the new potential difference. Plugging in the given values,
V' = (3000 V)(1 - κ)/κ = 1000 V
Solving for κ, we find:
κ = 2
Therefore, the new capacitance is:
C' = κC = 2(1.77 × 10−9 F) = 3.54 × 10−9 F
which is equivalent to 5.4 × 10^10. Therefore, the answer is C.

The Plates of a Parallel Plate Capacitor Each Have an Area