B.What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"
Solution:
A) A charge q under an electric field of intensity E will experience a force F equal to:
[tex]F=qE[/tex]
In our problem we have [tex]q=-2.5 nC=-2.5\cdot 10^{-9} C[/tex] and [tex]F=18 nN = 18 \cdot 10^{-9} N[/tex], so we can find the magnitude of the electric field:
[tex]E= \frac{F}{q}= \frac{18\cdot 10^{-9}N}{2.5\cdot 10^{-9}C}=7.2 V/m [/tex]
The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.
B) The proton charge is equal to
[tex]e=1.6\cdot 10^{-19} C[/tex]
Therefore, the magnitude of the force acting on the proton will be
[tex]F=eE=1.6\cdot 10^{-19} C \cdot 7.2 V/m=1.15 \cdot 10^{-18} N[/tex]
And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.
