An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. After the charge has moved to point B, 0.500 m to the right, it has kinetic energy 5.00×10−7 J . Part A If the electric potential at point A is +30.0 V, what is the electric potential at point B? Express your answer with the appropriate units. VB = nothing nothing Request Answer Part B What is the magnitude of the electric field? Express your answer with the appropriate units. E = nothing nothing

The electric potential at point B is 83.33 V

The magnitude of the electric field at point B is 166.67 N/C  

Explanation:

Given;

charge of an object, q = −6.00×10⁻⁹ C

Kinetic energy  = 5.00×10⁻⁷ J

Electric potential at point A = +30.0 V

To determine the electric potential at point B, we apply the following formula;

Fd = qEd

Also, Fd is work done in moving the charge to 0.5 m = K.E

Again, Ed is the electric potential at point B (VB)

Substitute for V and K.E in the above equation, we will have;

K.E = q(VB)

VB = K.E/q

To determine the magnitude of the electric field at B, we use equation below;

V = Ed

scientific laws are the result of testing theories and proving them to be facts.

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop