Question

if a proton is placed in a uniform electric field, what would be the magnitude and the direction of this field if the electrostatic force acting on the proton is just to balance it's weight

Answer 1

There are two forces acting on the proton, the electrostatic force and the weight; we also know that the forces are balanced which means that they have to be equal, then we have:

`\begin{gathered} F_e=W \\ \text{ where} \\ F_e\text{ is the electrostatic force} \\ W\text{ is the weight} \end{gathered}`

Now, we know that the electrostatic force is related to the electric field by:

`F_e=qE`

and that the weight is given by:

`W=mg`

Plugging these in the first equation we have:

`\begin{gathered} qE=mg \\ E=(mg)/(q) \end{gathered}`

The mass and charge of the proton are:

`\begin{gathered} m=1.67*10^(-27) \\ q=1.6*10^(-19) \end{gathered}`

plugging the values in the expression for the field we have:

`\begin{gathered} E=((1.67*10^(-27))(9.8))/(1.6*10^(-19)) \\ E=1.02*10^(-7) \end{gathered}`

Therefore, the magnitude of the field is:

`E=1.02*10^(-7)\text{ }(N)/(C)`

Since a proton has a positive charge it will move in the direction of the field; in this case we need the force to be pointing up since the weight points down.

Therefore, the direction of the field is up.