A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 3.90 and whose dielectric strength is 2.20  108 V/m. The desired capacitance is 0.200 µF, and the capacitor must withstand a maximum potential difference of 4.00 kV. Find the minimum area of the capacitor plates.

A capacitor is constructed out of two foil wrappers that have a length of 8.7 cm and width of 2.1 cm, and the plates (wrappers) are spaced 7 μm apart. If the capacitor is filled with a dielectric that has a constant of 120, what is the capacitance? (in F) a. 6.69×10-8 F  b. 3.61×10-6 F c.  2.31×10-9 F  d. 1.92×10-11 F e.  2.77×10-7 F

You have a capacitor made of two sheets of foil of area 2.7 cm2 separated by a distance 0.5 mm. a) what is the capacitance (in pF(10-12F)) of this capacitor? b) what is the voltage across the capacitor if 7 x 10-12 C is stored? c) how much energy is stored in the capacitor?

A capacitor of 3.23µF has an area of 6.35mm^2. Determine the separation distance between the two plates. If the new capacitance is now 5.63mF, what is the value of the dielectric inserted to the capacitor? What is the new voltage if the capacitor is connected to a 110 volts source? If the electric field created by two plates 8.99x10^4 N/C and a working voltage of 63 volts. Will the capacitor experience a dielectric breakdown? Please prove your answer. Note: This question is under the topic "Capacitance". Thank you!