A moving particle encounters an external electric field that decreases its kinetic energy from 9450 eV to 6990 eV as the particle moves from position A to position B. The electric potential at A is -43.0 V, and that at B is +39.0 V. Determine the charge of the particle. Include the algebraic sign (+ or -) with your answer. Number i Units Lower potential Higher potential VB E SUBROL

I Review A 13.0 nC charge is at x = Ocm and a -1.4 nC charge is at = 3.0 cm Part A At what point or points on the a-axis is the electric potential zero? Express your answer in centimeters. If there is more than one answer, give each answer separated by a comma. Vα ΑΣφ ? x0 = cm Submit Request Answer

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A proton with an initial speed of 850,000 m/s is brought to rest by an electric field. ▾ Part A Did the proton move into a region of higher potential or lower potential? O Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of higher potential to a region of lower potential. O Because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of higher potential to a region of lower potential. O Because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of lower potential to a region of higher potential. Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential. Submit ✓ Correct Here we learn how to determine the distribution of the electric potential based on the movement of a charged particle. Part B Previous Answers What was the potential difference…