• Support PF! Buy your school textbooks, materials and every day products Here!

Moving Proton in an Electric Field

  • Thread starter Fred91
  • Start date
  • #1
2
0
1.
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




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.
 

Answers and Replies

  • #2
gneill
Mentor
20,793
2,773
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).
Post your reasoning for your Gauss' Law derivation for the electric field near the plate.

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.
Voltage is not measured in eV. Voltage is measured in, well, Volts, which in turn can be expressed as Joules per Coulomb.

The eV is a unit of energy. It's the KE that would result from an electron (with unit fundamental charge 1.6*10^-19 Coulombs) falling through a potential difference of 1 volt (1 Joule per Coulomb). So you end up with the eV being 1.6*10^-19 Joules.

So when the problem gives you the kinetic energy of the proton in eV, it's not giving you a voltage but rather the energy in convenient units. The eV units are convenient for these types of problems because it immediately tells you what size of potential difference (volts) the unit charge must fall through in order to achieve a change in energy of the same size.

The electric potential difference in a uniform electric field (like the plate is providing) is simply the field magnitude multiplied by the change in distance parallel to the field lines.

In order to bring to rest a body with kinetic energy KE, an amount of work equal to that KE must be done on the body. You're given the KE of the proton in eV units...
 

Related Threads on Moving Proton in an Electric Field

Replies
1
Views
5K
Replies
5
Views
6K
Replies
2
Views
5K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
10
Views
11K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
8
Views
21K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
1
Views
2K
Top