A uniform electric field with a magnitude of 250 N/C is directed in the positive x direction. A 12 μC charge moves from the origin to the point (20.0 cm, 50.0 cm). What is the change in the electrical potential ener- gy of the system as a result of the change in position of this charge?

-0.6 mJ

Explanation:

We first find the change in electric potential ΔV from

ΔV = -∫E.dr = -Ercosθ were E = magnitude of electric field = 250 N/C and r = distance of charge from the origin = √[(20 - 0)² + (50 - 0)²] = √2900 = 53.58 cm = 0.536 m. The direction is θ = tan⁻¹(50/20) = 68.2 which is the angle between E and r.

So ΔV = -Ercosθ = - 250 N/C × 0.536 m × cos68.2 = -49.77 V

The electrical potential energy ΔU = qΔV were q = care = 12 μC = 12 × 10⁻⁶ C

ΔU = qΔV = 12 × 10⁻⁶ C × -49.77 V = - 597.2 × 10⁻⁶ J = - 0.5972 mJ ≅ -0.6 mJ

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

electrical power