# Solving Two Charged Plates Homework: Magnetism, Frequency & Charge Density

• scoutdjp2012
In summary, a proton is shot at high speed between two charged plates and is undeflected by the electric field. This indicates that the force exerted by the electric field is balanced by the Lorentz force due to a magnetic field into the paper. Using the definition of electric field and the Lorentz force law, the magnitude of the magnetic field can be found. The proton's trajectory will be a circle with a radius determined by the centripetal acceleration caused by the magnetic field. The time it takes to make a half circle and the period and frequency of the proton's motion can be easily calculated once the radius is known. Finally, the charge density for the parallel plate can be found using the formula for electric field in the interior of

## Homework Statement

A proton is shot at 5x10^6 m/s between two charged plates. When E between the plates is 15,000 N/C, the proton is undeflected by the field by a magnetic Field B into the paper
.
1) What is the magnitude of the magnetic field?
2) if the proton now enters a region with only the magnetic field, what direction and with what radius will it move?
3) How long will it take to make a half circle?
4) What would the period and frequency of the proton be?
5) find the charge density for the parallel plate

B = mv/qr
E=KQ/r^2

## The Attempt at a Solution

I figured out the magnitude of the magnetic field as 20.84 using B = mv/qr
for number 2, using the Right Hand Rule, the force is pointing upwards, and using E = KQ/r^2 i found the radius of the proton.

Can someone help with the last three?

Thread moved from Calculus section. For problems 3, 4, and 5, please show some effort. For problem 3, do you know how fast the proton is moving? In problem 2, you didn't find the radius of the proton -- you found the radius of the arc along which it is moving. Do these give you a hint?
See if you reason out at least a start for the three you're asking about.

scoutdjp2012 said:

## Homework Statement

A proton is shot at 5x10^6 m/s between two charged plates. When E between the plates is 15,000 N/C, the proton is undeflected by the field by a magnetic Field B into the paper
.
1) What is the magnitude of the magnetic field?
2) if the proton now enters a region with only the magnetic field, what direction and with what radius will it move?

## Homework Equations

B = mv/qr
E=KQ/r^2
What are the meanings of the letters m, v, q, r Q, k in the formulas?
Is E=KQ/r^ relevant here? The electric field is given between two charged plates.
scoutdjp2012 said:

## The Attempt at a Solution

I figured out the magnitude of the magnetic field as 20.84 using B = mv/qr
for number 2, using the Right Hand Rule, the force is pointing upwards, and using E = KQ/r^2 i found the radius of the proton.
How did you figure out the strength of the magnetic field with B = mv/qr? What is r in that formula? Was it given? What is the unit of B?
Again, what is r in the formula E = KQ/r^2? Is that formula relevant here?

As others here have noted, you don't seem to fully understand the meaning of ##r## in Coulomb's formula ##E = KQ / r^2##. The ##r## in this problem is something else. (Hint: Think centripetal acceleration.)

You've got another incorrect formula in part (1). Think simpler--the electric field exerts a force on the proton; so does the magnetic field. These forces are given by the definition of the electric field and the Lorentz force law, respectively. For the proton to be "undeflected," they must cancel out--so set them equal to each other and see what happens.

Once you find the proton trajectory's radius, parts 3 and 4 become trivial, as long as you remember the relationship between the period and frequency. Part (5) is just an application of a formula you don't seem to have remembered--the formula for the electric field in the interior of a parallel-plate capacitor. You can either look it up or derive it using Gauss's Law.