Generally, the two plates of a capacitor are oppositely charged. Right?lys04 said:Homework Statement: The electric field outside a conducting plate of charge is given by sigma/epsilon right? Then why not for a capacitor, since that is 2 conducting plates, is the electric field 2sigma/epsilon using superposition principle?
Relevant Equations: E=sigma/epsilon
The error in post #1 is much more fundamental than any concerns about fringing effects.Tom.G said:
The electric field between two parallel plates is a uniform field created by the potential difference (voltage) applied across the plates. It is calculated as the voltage difference divided by the distance between the plates, E = V/d, where E is the electric field, V is the voltage, and d is the separation distance.
The electric field strength (E) between parallel plates can be calculated using the formula E = V/d, where V is the voltage applied across the plates and d is the distance between them. The result is expressed in volts per meter (V/m).
The electric field between parallel plates is primarily affected by the voltage applied across the plates and the distance between them. Increasing the voltage increases the electric field strength, while increasing the distance between the plates decreases the electric field strength.
Yes, the electric field between two large, closely spaced, parallel plates is generally considered to be uniform across the region between the plates. This uniformity holds true as long as edge effects, which occur near the edges of the plates, are negligible.
The presence of a dielectric material between parallel plates reduces the electric field strength. This is because the dielectric material increases the capacitance of the plates, allowing them to store more charge for the same applied voltage. The electric field in the presence of a dielectric is given by E = V/(d * κ), where κ is the dielectric constant of the material.