Question

A positive charge of 5.0×10−4 C is in an electric field that exerts a force of 2.5×10−4 N on it. What is the magnitude of the electric field at the location of the charge?

Answer 1

Force experienced by a charge in an electric field is given by,

`F_(electric)=qE`

where q is the charge, E is the electric field,

Given,

`\begin{gathered} q=5.0*10^(-4)C \\ F_(electric)=2.5*10^(-4)N \end{gathered}`

Hence,

`\begin{gathered} 2.5*10^(-4)=5.0*10^(-4)* E \\ E=0.5\text{ N/C} \end{gathered}`

Result: 0.5 N/C