A uniform electric field of magnitude 277V/m is directly in the positive x-direction. A 1.33E-5C charge moves from the origin to the point (x,y)=(19.9cm, 46.0cm). What was the change in the potential energy of this charge?Through what potential difference did the charge move?

An ionized nitrogen molecule (N₂+) at point A has charge +e and moves at 1.92 × 10³ m/s in the positive x-direction. A constant electric force in the negative x-direction slows the molecule to a stop at point B, a distance of 0.796 mm past A on the > 2 axis. Calculate the x-component of the electric field and the potential difference between points A and B. (The mass of a nitrogen molecule is 4.65 x 107 kg and the fundamental charge is e = 1.60 × 10−¹9 C.) -26 HINT (a) the x-component of the electric field (in V/m) 672.96482 X V/m (b) the potential difference between points A and B (in V) V

The work done by an external force to move a -8.60 μC charge from point a to point b is 1.90x10-³ J. Part A If the charge was started from rest and had 4.77x10-4 J of kinetic energy when it reached point b, what must be the potential difference between a and b? ——| ΑΣΦ Vb - V₂ = ? V

A uniform electric field of magnitude 280 V/m is directed in the negative y direction as shown in the figure below. The coordinates of point A are (-0.350, -0.400) m, and the coordinates of point B are (0.250, 0.450) m. Calculate the electric potential difference VB − VA (in V) using the dashed-line path.