pat666 said:Homework Statement
The capacitance of a water-filled parallel-plate capacitor with plate area 3.0×10^−3m^2 and separation 4.0×10^−2 mm is
0.66 uF
1.3 nF
53 nF
660 nFHomework Equations
C=E_0A/d
The Attempt at a Solution
C=8.85*10^-12*3*10^-3/4*10^-5
C=6.64*10^-10
This is none of the options. I think that the constant 8.85*10^-12 may be wrong since the capacitor is water not air?
Simple capacitance refers to the ability of a system to store electric charge. It works by creating an electric field between two conductive plates separated by an insulating material, known as a dielectric. This field stores energy in the form of an electric charge, which can be released when needed.
The unit of capacitance is the farad (F), which is equal to one coulomb of charge per volt of potential difference. Capacitance can be measured by using a capacitance meter, or by applying a known voltage to the capacitor and measuring the resulting charge.
The capacitance of a system is affected by several factors, including the distance between the plates, the surface area of the plates, and the type of dielectric material used. Generally, a larger distance between plates or a smaller surface area will result in a lower capacitance, while a higher capacitance can be achieved by using a larger surface area and a higher dielectric constant material.
Capacitors have a wide range of practical applications, including energy storage, power factor correction, filtering, and timing circuits. They are also commonly used in electronic devices such as computers, televisions, and mobile phones.
The capacitance of a system can be calculated using the formula C = Q/V, where C is the capacitance in farads, Q is the charge in coulombs, and V is the potential difference in volts. Additionally, there are specific formulas for calculating the capacitance of different types of capacitors, such as parallel plate capacitors, cylindrical capacitors, and spherical capacitors.