a proton is released from rest at a positivity charged plate that is in a parallel...

a proton is released from rest at a positivity charged plate that is in a parallel plate capacitor . it hits the negative charged plate after 2.54 x10^-6 s. the electric field is 134 N/C between the plates

1) find the magnititude of the surface charge density at each plate

2) what is the magnitude of force that the proton feels moving through plates

3)what is the proton’s final speed as it hits the negative plate

4)what is the potential difference between the two plates

Expert Solution

Initially, when the proton was released from the rest at positive plate, proton's electrostatic potential energy increases and kinetic energy decreases. By the time it reaches positive plate it comes to rest momentarily. Proton now, faces attractive force from negatively charged plate of capacitor. Proton travels under the influence of attractive force towards negatively charged plate and its kinetic energy increases and potential enegy decreases.

genius_generous answered 2 years ago

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