5. A charge moves a distance of 2.0 cm in the direction of a uniform electric field having a magnitude of 215 N/C. The electrical potential energy of the charge decreases by 6.9 x10-19 J as it moves. Find the magnitude of the charge on the moving particle. (Hint: The electrical potential energy depends on the distance moved in the direction of the field.)

The magnitude of charge on the particle is .        

Explanation:

Given that,

The magnitude of electric field, E = 215 N/C

The electrical potential energy of the charge decreases by, as it moves.

We need to find the magnitude of the charge on the moving particle. The change in electric potential energy is given by :

.........(1)

The electric potential in terms of electric field is given by :

Equation (1) becomes :

So, the magnitude of charge on the particle is .                                    

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

12000j is work done