If the electron experiences a force
of 5.6 x 10-6 N, how large is the field between the
deflection plates?
To determine the field between the deflection plates, we can use the formula for the force experienced by a charged particle in an electric field.
The formula is given by:
F = q * E
Where F is the force experienced by the particle, q is the charge of the particle, and E is the electric field strength.
In this case, we know the force experienced by the electron (F = 5.6 x 10^(-6) N), and we know that the charge of an electron is 1.6 x 10^(-19) C. We can rearrange the formula and solve for E:
E = F / q
Plugging in the known values, we get:
E = (5.6 x 10^(-6) N) / (1.6 x 10^(-19) C)
Now, let's simplify the expression:
E = (5.6 / 1.6) * (10^(-6) / 10^(-19))
E = 3.5 x 10^13 N/C
Therefore, the magnitude of the electric field between the deflection plates is 3.5 x 10^13 N/C.