Answer to Question #112107 in Electric Circuits for asdasd

Question

what are the magnitude and direction of the electric field at point midway between a -8.0μC and a 7.00μC charge 8.0 cm apart? assume no other charges are nearby

Answer 1

$|\vec E|=|\vec E_1|+|\vec E_2|=\frac{1}{4\pi \epsilon_0}(\frac{|q_1|}{(r/2)^2}+\frac{q_2}{(r/2)^2})$ or

$|\vec E|=\frac{1}{\pi\epsilon_0r^2}(|q_1|+q_2)\approx8.4\sdot10^7 N/C$ (Newton per coulomb), where $1/4\pi\epsilon_0\approx9\sdot10^9Nm^2c^{-2},r=8\sdot10^{-2}m$

$|q_1|=8\sdot10^{-6}C,q_2=7\sdot10^{-6}C$

The direction of the field vector towards the negative charge