Question

A charged particle moves from point A to point B in an external electric field, and in the process its kinetic energy decreases fro, 87.6 J at A to 57.3 J at B. The electric potential at A is -48.0 V, and the electric potential at B is 18.0 V. What is the charge of the particle, including sign

Answer 1

-0.46 C

Step-by-step explanation:

The relationship between total kinetic energy, KE, and total electric potential, V is:

ΔKE = ΔV * q

where ΔKE = change in kinetic

ΔV = change in voltage

q = charge

The kinetic energy changes from 87.6 J to 57.3 J while the electric potential changes from -48.0 V to 18.0 V.

Therefore:

57.3 - 87.6 = (18.0 - (-48.0)) * q

-30.3 = 66q

=> q = -30.3 / 66

q = -0.46 C

The charge of the particle is -0.46 C.