In which direction is the electric field on the cylindrical Gaussian surface? Check all that apply. Check all that apply. perpendicular to the curved wall of the cylindrical Gaussian surface tangential to the curved wall of the cylindrical Gaussian surface perpendicular to the flat end caps of the cylindrical Gaussian surface tangential to the flat end caps of the cylindrical Gaussian surface

Ithink it is letter b, i may be wrong but i think that is the answer

It really doesn't matter for any of the questions that you've listed.

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

Maybe because they hurt or you bleed a lot