Question

A + 10 nC point charge and a - 3.62 nC point charge are 4.41 cm apart. What is the electric field strength at the midpoint between the two charges?

Answer 1

ThereExplanation:

We can represent the situation with the following figure

Therefore, we need to calculate the electric field generated by each charge. The electric field can be calculated as

$E=\frac{kq}{r^2}$

Where k = 9 x 10^9 N m²/C², q is the charge, and r is the distance.

For the first charge, we need to replace q = 10 nC = 10 x 10^(-9) C and r = 2.205 cm = 0.02205 m, then

$E_1=\frac{(9\times10^9)(10\times10^{-9})}{(0.02205)^2}=1.8\times10^5\text{ N/C}$

In the same way, we can calculate the electric field for the second charge, replacing q = 3.62nC = 3.62 x 10^(-9) C and r = 0.02205 m

$E_2=\frac{(9\times10^9)(3.62\times10^{-9})}{(0.02205)^2}=0.67\times10^5\text{ N/m}$

Then, the electric field strength at the midpoint of the two charges is

$\begin{gathered} E=E_1+E_2 \\ E=(1.8\times10^5)+(0.67\times10^5) \\ E=2.47\times10^5\text{ N/C} \end{gathered}$

Therefore, the answer is 2.47 x 10^(5) N/C