Answer with Explanation:
We know that by Gauss's Law
[tex]E\cdot Flux=\frac{q_{inside}}{\epsilon _o}[/tex]
Where
[tex]\epsilon _o[/tex] is permitivity of free space
[tex]q_{inside}[/tex] is the net charge in the hull
Part a)
Applying the given values we get
[tex]E\cdot Flux=\frac{(2.75+27-9-43.5)\times 10^{-6}C}{8.85\times 10^{-12}}\\\\\therefore E\cdot flux=-2.5\times 10^{6}[/tex]
Part b)
Since the sign of the electric flux is negative we conclude that the electric field lines entering the submarine are more than the electric field lines leaving the submarine. Since negative sign implies that flux by negative charges is dominant.
