There are two plates.Jrlinton said:Homework Statement
Two large metal plates of area 1.1 m^2 face each other, 6.2 cm apart, with equal charge magnitudes but opposite signs. The field magnitude E between them (neglect fringing) is 50 N/C. Find |q|.Homework Equations
|q|=σA
E=σ/(2Eo)⇒σ=2EoE
|q|=2EoEA
The Attempt at a Solution
I thought that after I did the rearranging of formulas above that this would be a pretty simple problem
|q|=2EoEA
|q|=2(8.99E-9C)(50N/C)(1.1m^2)
|q|=9.889E-7C
The electric field between two metal plates is directly proportional to the charge on the plates. This means that as the charge on the plates increases, the electric field between them also increases.
The distance between two metal plates does not directly affect the charge on the plates. However, it does affect the electric field between the plates, which in turn affects the force experienced by the charges on the plates.
The material of the plates does not have a significant impact on the charge between them. However, different materials may have different conductivity, which can affect how easily charges can move between the plates.
Dielectric material, such as air or plastic, can affect the charge between two metal plates. When a dielectric material is placed between the plates, it can decrease the electric field and increase the capacitance of the system.
The voltage difference between two metal plates is directly proportional to the charge on the plates. This means that as the voltage difference increases, the charge on the plates also increases. This relationship is known as capacitance and is described by the equation Q = CV, where Q is charge, C is capacitance, and V is voltage.