Question

An ion of net charge -2 is accelerated through a potential difference of 250 volts. What is the magnitude of the force on the ion from the electric field between the plates if the plates are 3.4 cm apart?

Answer 1

Answer:

$2.35\cdot 10^{-15} N$

Explanation:

When a charge is in an electric field, it experiences a force given by

$F=qE$

where

q is the charge

E is the strength of the electric field

The field between two parallel plates is uniform, so it can be rewritten as

$E=\frac{V}{d}$

where

V is the potential difference between the plates

d is the separation between the plates

So the first equation becomes

$F=\frac{qV}{d}$

In this problem we have:

$q=2e=2(1.6\cdot 10^{-19}C)=3.2\cdot 10^{-19}C$ is the magnitude of the charge

V = 250 V is the potential difference between the plates

d = 3.4 cm = 0.034 m is the separation between the plates

So the magnitude of the force is

$F=\frac{(3.2\cdot 10^{-19})(250)}{0.034}=2.35\cdot 10^{-15} N$