Question

The weight of a proton is 1.64×10−26 N. The charge on a proton is +1.60×10−19 C. If a proton is placed in a uniform electric field so that the electric force on the proton just balances its weight, what is the magnitude and direction of the field?

Answer 1

Given:

The weight of the proton is: W = 1.6 × 10^(-26) N.

The charge on a proton is: q = 1.60 × 10^(-19) C

To find:

The magnitude and the direction of the electric field.

Step-by-step explanation:

The weight of the proton is the force that the proton experiences due to its mass and acceleration. The electric force balances the weight of the proton. Thus we have,

F = W

Here, F is the electric force a proton experiences when it is placed in an electric field and W is the weight of the proton,

The force experienced by a photon when it is placed in an electric field is given as,

`F=Eq`

Here, E is the electric field.

Rearranging the above equation and substituting the values, we get:

`\begin{gathered} E=(F)/(q) \\ \\ E=\frac{1.64*10^(-26)\text{ N}}{1.60*10^(-19)\text{ C}} \\ \\ E=1.025*10^(-7)\text{ N/C} \end{gathered}`

Thus, the magnitude of the electric field is 1.025 × 10^(-7) N/C.

The charge on the proton is positive and when it is placed in the electric field, the electric force on the proton is balanced by the weight of the proton. Thus, The direction of the electric force is opposite to the direction of the weight of the proton which is radially outward.

Final answer:

The magnitude of the electric field is 1.025 × 10^(-7) N/C and it has a radially outward direction that is opposite to the wight of the proton.