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?

Answers

Answer 1

Answer:

The magnetic field strength will allow the electron to pass between the plates without being deflected is 0.0033 T.

Electric field strength

The electric field strength of the electron is calculated as follows;

E = V/d

E = 200/(0.01)

E = 20,000 V/m

Magnetic field strength

The magnetic field strength is related to electric field in the following formula;

qvB = qE

vB = E

B = E/v

B = (20,000)/(6 x 10⁶)

B = 0.0033 T

Thus, the magnetic field strength will allow the electron to pass between the plates without being deflected is 0.0033 T.

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Answer 2

Answer:

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

$E=(\Delta V)/(d)=(200 V)/(0.01 m)=2\cdot 10^4 V/m$

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:

$F_E = F_B\nqE=qvB$

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:

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

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Answers

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= 600 / 3
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Explanation:

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Answers

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