Question

Two point charges are separated by 10.0 cm and have charges of +2.00 µC and -2.00 µC, respectively. What is the electric field at a point midway between the two charges? (ke = 8.99 x10^9 N·m^2/C^2)

Answer 1

Let's write the electric field at a point midway due to +2.00 µC charge:

$$E_1=\dfrac{kq_1}{r^2}=\dfrac{8.99\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot2\cdot10^{-6}\ C}{(0.05\ m)^2}=7.192\cdot10^{-6}\ \dfrac{N}{C}.$$

Let's write the electric field at a point midway due to -2.00 µC charge:

$$E_2=\dfrac{kq_1}{r^2}=\dfrac{8.99\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(-2\cdot10^{-6}\ C)}{(0.05\ m)^2}=-7.192\cdot10^{-6}\ \dfrac{N}{C}.$$

Finally, we can find the net electric field at a point midway between the two charges:

$$E_{net}=E_1-E_2,$$ $$E_{net}=7.192\cdot10^{-6}\ \dfrac{N}{C}-(-7.192\cdot10^{-6}\ \dfrac{N}{C})=14.384\cdot10^{-6}\ \dfrac{N}{C}.$$