Question

A small particle has charge -2.40 μc and mass 2.30×10−4 kg . it moves from point a , where the electric potential is va = 300 v , to point b , where the electric potential vb = 770 v is greater than the potential at point a . the electric force is the only force acting on the particle. the particle has a speed of 5.50 m/s at point a .

Answer 1

The potential difference (voltage) through which the electron traveled is approximately -360.83 V.

Step-by-step explanation:

To find the potential difference (voltage) through which the electron traveled, we can make use of the equation:

V = KE/e

where V is the potential difference, KE is the kinetic energy of the electron, and e is the charge of the electron.

First, calculate the kinetic energy of the electron:

KE = 1/2 * m * v^2

where m is the mass of the electron and v is its speed.

Substituting the given values into the equation, we have:

KE = 1/2 * (2.30x10^-4 kg) * (5.50 m/s)^2

Solving for KE, we get:

KE = 1/2 * (2.30x10^-4 kg) * (5.50 m/s)^2 = 8.66x10^-4 J

Next, substitute the values of KE and e into the first equation to calculate the potential difference (voltage):

V = KE/e = (8.66x10^-4 J) / (-2.40x10^-6 C) = -360.83 V

So, the potential difference through which the electron traveled is -360.83 V.