\begin{tabular}{|l} Question 8 \\ A 5 mC charge is in a \( 25 \mathrm{~N} / \mathrm{C} \) electric field. What is the force on the charge? \\ \hline\end{tabular}

To find the force acting on a charge in an electric field, you can use the formula: \( F = qE \), where \( F \) is the force, \( q \) is the charge, and \( E \) is the electric field strength. In this case, with a charge of 5 mC (which is \( 5 \times 10^{-3} \) C) and an electric field of \( 25 \, \mathrm{N/C} \), you would calculate: \( F = (5 \times 10^{-3} \, \mathrm{C}) \times (25 \, \mathrm{N/C}) = 0.125 \, \mathrm{N} \). So, the force on the charge is 0.125 N! Now, imagine this in action! If you place that tiny 5 mC charge in that electric field, it feels a little nudge, pushing it away as if the field is saying, "Hey, move that way!" It’s like having a friendly shove when you're trying to cross the street—just a little help from the physics of electric fields!