Physics question: Magnetic Force problem?

Question

I'm having trouble solving this question, any help would be appreciated:

a) Two parallel plates having charges of equal magnitude but opposite sign are separated by 10.1 cm. Each plate has a surface charge density of 39.9 nC/ m2. A proton is released from rest at the positive plate. Determine the potential difference between the plates.
b) What is the energy of the proton when it reaches the negative plate.
c) What is the speed of the proton just before it strikes the negative plate.

If you someone could give me a step by step of how to solve this I'd really apprecaite it.

Thanks

Whats A in your top equation?

Answer 1

you can consider this as a parallel plate capacitor
Electric field inside capacitor E = sigma/eo (constant inside)
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sigma= Q/A = charge/area and
eo = permittivity of free space = 1/(4pi*9*10^9)
potential difference between plates = dV = (V+) - (V-)
dV = - E dx
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negative sign just indicates that when x is increasing towards (-ve) plate, potential V is decreasing towards -ve plate or V+ > V-
so we will use (- x) to balance sign paraity.
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dV = + E* x = sigma * x / eo
dV = (4*3.14*9*10^9)*39.9*10^-9*0.101 = 455.54 Volts

How can we bring proton near the +ve plate, and then release it from rest (u=0)??
Net electric field just outside +ve plate = 0 >>> only when 2 plates are infinite planes or separation between these 2 plates is very very small in comparison to area (A) of plates.

10.1 cm separation does not give any information as size of plates is kept unknown (or whether 10.1 cm is small enough to consider E (outside) =0

also, given that E=0 outside (explanation nedded) and proton can be readily swung in at +ve plate, till you hold it at u=0 and release it (it would have aquired some potential energy) from the work done by you either in holding or bringing it from infinity.

problem lies in boundary condition at +ve plate. once it has to be take as x=0 for kinetic energy (PD) and it has to be also taken as x = - infinity to facilate existence of proton around it.

The problem can be solved by taking +ve plate at V+ = 0 (an infinity of potential refernce system) which can be seen as +ve plate earthed. >>> this is the only way you can conjure up a proton there and release it >> then cut out the earthing

problem is compounded by the 10.1 cm region of arrangement >> bcause inside E = constant and inverse square variation does not hold due to infinite planes.

you provide the latent info if any.

Answer 2

THis is not magnetism, it is electrostatics. Okay each plate has a surface charge density, call that s. The electric field between the plates (far from the edges) is found using Gauss' theorem,

E = charge* surface area/plate separation = e0*s*A/z, e0 = permittivity of free space

The potential is V = Ez = e0*s*A

Th energy gained by a proton moving from one plate to the next is Ep = q*V where q =1.6*10^-19 Coul.

Speed of the proton, v is given by conservation of energy:

qV = 1/2*mv^2 ---> v = sqrt(2qV/m) = sqrt(2q*e0*s*A/m)
where m = 1.6*10^-28 kg = mass of proton

Answer 3

I acquired fifty eight.8m I did this through figuring out the pace at which it might hit the plate (6 x nine.eight) (taking nine.eight to be the acceleration because of gravity). This offers a pace of fifty eight.eight m/s. a million/three of that is 19.6 so that is the speed of the ball after hitting the plate. The ball falls for two seconds, so I figure out the pace it's traveling after simplest a million moment. Since this may increasingly supply the ordinary pace of the ball during the ones two seconds. so 19.6+nine.eight = 29.four m/s So count on the ball is traveling at an ordinary pace of 29.4m/s for two seconds, then the gap it's going to have travelled is the time x pace. two (s) x 29.four (m/s) = fifty eight.eight (m) . So the plate is fifty eight.eight m from the bottom. P.S: I could not discover my calculator so that is the satisfactory I would do utilising "intellectual maths" as I could not keep in mind the right equation. It works logically regardless that!