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If I have a spherical shell with a uniform positive charge desnity on its surface is near a positive point charge q (sitting in quadrant II area)

In consideration of the electric field strength and point p, why must I consider the principle of superposition.

If E1 in the shell in 0, how can the electric field be greater then 0?, why does it point in the north-west direction?

Thanks

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    $\begingroup$ The third line in red explains this. $\endgroup$ – G. Smith Mar 17 at 23:38
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$E_1$ is the E-field due to the shell, so it would be $0$ by gauss' law. But the particle also creates a nonzero E-field ($E_2$). This electric field is entirely responsible for the nonzero E-field inside the shell.

Superposition here is just adding up the E-fields at point P $$E_1 (=0) +E_2=E_2$$

It's going to be NW because that is the direction of the only field ($E_2$) that is actually making a contribution at that point (if you were to take a straight edge and draw a line pointing from the point charge to P, you'd find it pointing in the NW direction).

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