Question

The electric force experienced by a charge of 1.0 × 10−6 C is 1.5 × 10−3 N. Find the magnitude of the electric field at the position of the charge.

Answer 1

The magnitude of the electric field at the position of the charge is, 1500 N / C .

Explanation:

Step 1:

Given values in the question :

The electric force with a charge of $1.0 \times 10^{-6} \mathrm{C} \text { is } 1.5 \times 10^{-3} \mathrm{N}$

            F is the electric force  

            q is the charge

            $\mathrm{F}=1.5 \times 10^{-3} \mathrm{N}$

             $q=1.0 \times 10^{-6} \mathrm{C}$

Step 2:

        Force = charge × electric field

               F = q E

               $E=\frac{E}{q}$

               $E=\frac{15 \times 10^{-3} \mathrm{N}}{1.0 \times 10^{-6}}$

               $E=\frac{1.5 \times 10^{3} \mathrm{N}}{1.0 \mathrm{C}}$

               $E=1.5 \times 10^{3} \mathrm{N} / \mathrm{C}$

               $E=1500 \mathrm{N} / \mathrm{C}$

Thus, the electrical field at the charging position is 1500 N / C .