A point charge of -3.90 nCis located at the origin and a second charge of 5.30 n

Question

A point charge of -3.90 nCis located at the origin and a second charge of 5.30 nC is located onthe x axis at x = 1.50 m.Calculate the electric flux through a sphere centered at the originwith radius 1.00 m.
I understandthat an equation suchas =Qenc/0 isused. If bothpoints are on the inside you can add them for Qenc. But when one is on the outside, do I subtract? Or maybeeven ignore the outside force?
I triedsubstracting but I got the wrong answer, so im thinking theresanother approach.
I understandthat an equation suchas =Qenc/0 isused. If bothpoints are on the inside you can add them for Qenc. But when one is on the outside, do I subtract? Or maybeeven ignore the outside force?
I triedsubstracting but I got the wrong answer, so im thinking theresanother approach.

Explanation / Answer

According to Gauss Law , Flux through a closed surface isequal to ( 1/ ) times charge enclosed with in thesurafce. So, flux through a sphere = Charge enclosed by sphere /                                     =q / where = permitivity of free space = 8.85 * 10 ^-12 C ^2 / N m^2           q =-3.9 * 10 ^ -9 C plug the values weget = 440.67 N m^ 2 / C

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