I don't know why the answer is wrong, I got K(final) = 1.1072x10^-15 J and K(initial) = 1.4576x10^-15J, with the work energy theorm I got q=(Kf-Ki)/(Vi-Vf) = +5.48x10^-18 C
I don't know why the answer is wrong, I got K(final) = 1.1072x10^-15 J and K(initial)...
Your answer is partially correct. Try again. A moving particle encounters an external electric field that decreases its kinetic energy from 9040 eV to 7610 eV as the particle moves from position A to position B. The electric potential at A is -48.0 V, and that at B is +14.0 V. Determine the charge of the particle. Include the algebraic sign (+ or -) wth your answer. Higher potential Lower potential VA 36.9 Number Units the tolerance is +/-2%
Learning Goal: To analyze an RC circuit to determine the initial voltage across a capacitor, the time constant, and the expression for the natural response of the capacitor voltage, and then to find other circuit quantities such as current,voltage, power, or energy. The natural response of an RC circuit is the response of the capacitor voltage to the sudden removal of a DC source. When this occurs, the capacitor releases its stored energy Figure < 10121〉 t 0 V. Figure...