The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, increasing its kinetic energy by 3.20×10−19 J . In what direction and through what potential difference Vb−Va does the particle move?a. The particle moves to the left through a potential difference of Vb-Va = 0.500 V. b. The particle moves to the left through a potential difference of Vb-Va = -0.500 Vc. The particle moves to the right through a potential difference of Vb-Va = 0.500 V. d. The particle moves to the right through a potential difference of Vb-Va = -0.500 V. e. The particle moves to the left through a potential difference of Vb-Va = 5.00 V. f. The particle moves to the right through a potential difference of Vb-Va = -5.00 V.

In which scenario is the least amount of work done on a crate? a force of 120 n moves it 25 m. a force of 115 n moves it 26 m. a force of 130 n moves it 22 m. a force of 125 n moves it 27 m.

An isolated conducting spherical shell carries a positive charge. part a which statement (or statements) about the electric field and the electric potential inside and outside the spherical shell is correct? which statement (or statements) about the electric field and the electric potential inside and outside the spherical shell is correct? electric potential inside the shell is constant and outside the shell is changing as 1/r2 both the electric potential and the electric field does change with r inside and outside the spherical shell electric potential inside and outside the shell is constant, but not zero electric potential inside the shell is constant and outside the shell is equal to zero electric field inside and outside the shell is constant (does not change with the position r), but is not equal to zero electric field inside and outside the shell is changing as 1/r (where r is the distance from the center of the sphere) electric field inside is equal to zero and outside the shell is constant, but not zero electric potential inside the shell is constant and outside the shell is changing as 1/r electric field inside and outside the shell is changing as 1/r2 electric field inside is equal to zero and outside the shell is changing as 1/r2 electric field inside and outside the shell is zero electric field inside is constant and outside the shell is changing as 1/r

When using the scientist method, which step comes last?