Question

An electron travels with speed 6.0 106m/s between the two parallel charged plates shown in the figure. The plates are separated by 1.0 cm and are charged by a200 V battery.What magnetic field strength will allow the electron to pass between the plates without being deflected?

Answer 1

Answer:

3.3 mT

Explanation:

First of all, we need to find the strength of the electric field between the two parallel plates.

We have:

$\Delta V=200 V$ (potential difference between the two plates)

$d=1.0 cm=0.01 m$ (distance between the plates)

So, the electric field is given by

$E=\frac{\Delta V}{d}=\frac{200 V}{0.01 m}=2\cdot 10^4 V/m$

Now we want the electron to pass between the plates without being deflected; this means that the electric force and the magnetic force on the electron must be equal:

$F_E = F_B\\qE=qvB$

where

q is the electron charge

E is the electric field strength

v is the electron's speed

B is the magnetic field strength

In this case, we know the speed of the electron: $v=6.0\cdot 10^6 m/s$, so we can solve the formula to find B, the magnetic field strength:

$B=\frac{E}{v}=\frac{2\cdot 10^4 V/m}{6.0\cdot 10^6 m/s}=0.0033 T=3.3 mT$