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 energy5.00×10−7 J .

A. If the electric potential at point A is +30.0 V, what is the electric potential at point B?

Answer 1

The electric potential at point B is +35.0 V.

Step-by-step explanation:

The electric potential at point B can be calculated using the formula:

VB - VA = -ΔE/q

Where VB is the electric potential at point B, VA is the electric potential at point A, ΔE is the change in electric potential energy (kinetic energy), and q is the charge. Rearranging the equation gives:

VB = VA - (ΔE/q)

Plugging in the given values:

• VA = +30.0 V
• ΔE = 5.00×10-7 J
• q = -6.00×10-9 C

Substituting these values into the equation gives:

VB = +30.0 V - (5.00×10-7 J)/(-6.00×10-9 C)

Solving the equation gives:

VB = +35.0 V