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

`2.35\cdot 10^(-15) N`

Step-by-step 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=(V)/(d)`

where

V is the potential difference between the plates

d is the separation between the plates

So the first equation becomes

`F=(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=((3.2\cdot 10^(-19))(250))/(0.034)=2.35\cdot 10^(-15) N`