The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
where
is the charge density
is the vacuum permittivity
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
Answer:
A. Light acts as particles, causing electrons on the surface it strikes to be destroyed.
Explanation:
- The photoelectric effect is a phenomenon that occurs when light shined onto a metal surface causes the ejection of electrons from that metal. It was observed that only certain frequencies of light are able to cause the ejection of electrons.
If a battery with a potential difference of 1.5 volts is placed across the plates, the maximum capacitor will have a charge of 36 V.
<h3>What possible variations are there in a 1.5 volt battery?</h3>
1 V is, by definition, a potential energy differential between two places equal to one joule for every coulomb of charge. Your query is resolved by that. Between the sites where that potential difference is measured, 1.5V denotes a potential energy differential of 1.5 joules per coulomb.
<h3>How do you determine the difference in potential energy?</h3>
ΔV=VB−VA=ΔPEq. By dividing the potential energy of a charge q that has been transported from point A to point B by the charge, we may define the potential difference between points A and B as VBVA. The joules per coulomb, sometimes known as volts (V) in honor of Alessandro Volta, are the units of potential difference.
To know more about potential energy difference visit ;
brainly.com/question/12807194?
#SPJ4
their leaders can run for office
