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increase the capacitance of the capacitor by a factor of two. This is because capacitance is directly proportional to the area of the plates.
by using capacitor plates. The length,area ,thickness and type of the plate determines the amount of charge a capacitor can store.
Pursuant to Ohms Law, we can deduce that the answer is the square root of Pi divided by C*R+A.
Yes. Increasing the plate area of a capacitor increases the capacitance. The equation of a simple plate capacitor is ...C = ere0(A/D)... where C is capacitance, er is dielectric constant (about 1, for a vacuum), e0 is electric constant (about 8.854 x 10-12 F m-1), A is area of overlap, and D is distance between the plates. (This is only a good estimate if D is small in comparison to A.) Looking at this, you can see that capacitance is proportional to plate area.
The basic geometry of a parallel plate capacitor does not affect its capacitance because capacitance is determined by the area of the plates and the distance between them, not their shape or size.
If the area of one plate of a parallel plate capacitor is increased while keeping the separation between the plates constant, the capacitance of the capacitor will increase. Capacitance is directly proportional to the area of the plates, as described by the formula ( C = \frac{\varepsilon A}{d} ), where ( C ) is capacitance, ( \varepsilon ) is the permittivity of the dielectric material between the plates, ( A ) is the area of the plates, and ( d ) is the separation distance. Thus, a larger plate area allows for greater charge storage, resulting in higher capacitance.
In a parallel plate capacitor, the second plate serves to create an electric field between the two plates when a voltage is applied. This configuration allows the capacitor to store electrical energy in the electric field created between the plates. The separation and area of the plates, along with the dielectric material (if present), determine the capacitor's capacitance, which indicates its ability to store charge. Essentially, the second plate works in conjunction with the first plate to facilitate charge separation and energy storage.
Area = pi*52 = 78.53981634 square inches
3.42*10^-11 farad.
You need more information than this to find your solution. There will be a a value for the electric field at which point the air will break down. you also need the area of one of the capacitors and the capacitence
no, because area of 1 fared capacitor is equal to double area of earth.
Magic. Look up capacitors on wikipedia!!A capacitor stores electrical charges in its plates.Both wrong. A capacitor stores energy as an electric field developed in the dielectric between its plates. A good dielectric with high permittivity (once called dielectric constant) concentrates this field, allowing more energy to be stored in a capacitor having the same plate area and separation but a dielectric of lower permittivity.