Question

• Two point charges, one +400.0 nC and the other -400.0 nC, located 20.00cm to the right of the first , are in vacuum. Determine the electric field (magnitude and direction) at a point midway between the charges.

Answer 1

$q_1=+400\space nC=400\times10^{-9}\space C$

$q_2=-400\space nC=-400\times10^{-9}\space C$

$E_1=\dfrac{kq_1}{r^2}=\dfrac{9\times 10^9\times 400\times 10^{-9}}{(10\times10^{-2})^2}$

$E_1=3.6\times10^5\space N/C\space(towards \space right)$

$E_2=\dfrac{kq_2}{r^2}=\dfrac{9\times 10^9\times 400\times 10^{-9}}{(10\times10^{-2})^2}$

$E_2=3.6\times10^5\space N/C\space(towards \space right)$

$E_{res}=E_1+E_2=7.2\times10^9\space N/C\space(towards \space right)$