What Magnetic Field Strength Keeps an Electron on a Straight Path?

AI Thread Summary
To determine the magnetic field strength that keeps an electron on a straight path between two parallel plates, the problem involves calculating the velocity of the electron after being accelerated through a potential difference of 1.7 kV. The kinetic energy gained by the electron can be expressed using the equation qU = (1/2)mv^2, allowing the calculation of its velocity, which is found to be 24,449 m/s. The relationship between the electric field and magnetic field is established using the equations F = qE and F = qvB, leading to vB = E. However, there seems to be a discrepancy in the calculations, particularly when substituting values for the electric field. The user is uncertain about the accuracy of their math, indicating a need for further verification.
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Homework Statement



In the figure, an electron accelerated from rest through potential difference 1.7 kV enters the gap between two parallel plates having separation 20.0 mm and potential difference 110 V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. What uniform magnetic field allows the electron to travel in a straight line in the gap?


Homework Equations


F=qvB
q=charge
v=velocity
B=magnetic field




The Attempt at a Solution



I honestly don't know where to even begin, I know I've got to be missing a formula because distance has got to be used somewhere in there, I'm just interested in getting an idea how to start this problem.
 
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If the electron travels in a straight line it doesn't accelerate upwards/downwards, which essentially means that the sum of all forces acting in the direction of the y-axis is zero. The magnetic force has to be equal, but opposite to that of the electric force.
 
Ok, so that means that F=-F, and i found another formula qvB=(mv^2)/r, but I'm sure that I need to find the velocity first and then use that and substitute and since i have two equations i would be able to solve for F which would allow me to find B, but I'm not sure how to find the velocity by only knowing the potential differences.
 
Forget qvB=(mv^2)/r. This equation describes an electron (or any other particle with a charge q) moving in a circular path, this is not true in your assignment.

If a particle with the charge q is accelerated with the voltage U it will gain kinetic energy according to

qU=\frac{1}{2}mv^2

From this equation you can solve the speed of the electron.
 
Ok so, I found the velocity to be 24,449 m/s and then by combining the equations qvB=F and F=qE I got the equation vB=E so 24449 m/s x B = E and E=V(volts) / d

so I plugged in numbers and 1.27e-10 but wrong... Is my math wrong?
 
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