Question

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

Answer 1

Answer:

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

$V_B = \frac{5*10^{-7}}{6*10^{-9}} = 83.33 \ V_B$

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

V = Ed

$E = \frac{V}{d} = \frac{83.33}{0.5} = 166.67 \ N/C$