- #1

Fred91

- 2

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A) A proton with kinetic energy of 4.95 × 106 eV is fired perpendicular to the face of a large plate that has a uniform charge density of σ = +8.10 μC/m2. What is the magnitude of the force on the proton?

B) How much work must the electric field do on the proton to bring it to rest?

2.

**E = σ/epsilon**

V=K*q/r

F= E*q

V= -E*r

W=q*v

V=K*q/r

F= E*q

V= -E*r

W=q*v

When I started making this thread I didn't understand how to do any of this, then while putting everything together I figured out how to do all of it, but I still have questions about the concepts. Particularly part B. I mostly understand part A.

3.

Part A) Since an elementary charge is 1.6*10^-19 I used this charge to figure out the magnitude of the force. Apparently somehow guass's law comes into play for part A making the formula needed become qσ/(2epsilon) If someone could explain to me why this is so I would appreciate it, when I did it without looking seeking help I only made the equation qσ/(epsilon).

Part B) I'm very confused about this method. I thought maybe if I took eV/q that would give me delta(V) which = delta(U)/q so eV=work, but apparently you can use work = q*voltage where the voltage is *already* measured in eV? If someone could explain to me I would appreciate it since the formula isn't in my book and I can't seem to find anywhere to derive it. Closest I can find is delta(v)=delta(U)/q where U is the Electric potential difference. This is the part of this problem I need explained. My book says q*delta(V) gives you eV. But yet to get the answer for this problem I did q*V, where V= 4.95*10^6eV and q = eV measured as 1.6*10^-19 J

So essentially I multiplied the energy given in eV by eV to get the work (which my book leads me to believe is actually how you change a value into eV)? Maybe I'm misunderstanding my book.