The Electric Potential Equation

In summary, an object with a charge of -4.00x10^-9 C is released from rest at point A in a uniform electric field and moves to point B, 0.500 m to the right, with a kinetic energy of 5.00x10^-7 J. The question asks for the magnitude of the electric field. The relevant equations used in the attempt at a solution are U = kQ/r^2, F = Eq, W = Fd, and U = Eqd. The poster also mentions setting the voltage difference equal to the electric field x charge x distance, but it is not clear how this is related to the solution.
  • #1
meaghan
15
1

Homework Statement



An object with charge q = −4.00×10−9 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 .What is the magnitude of the electric field?

Homework Equations


U = kQ/r^2
F = Eq
W = Fd
U = Eqd

The Attempt at a Solution


I was thinking that I could maybe set the voltage difference equal to the electric field * charge * distance since the part before asked that if the electric potential at point A is +30.0 V, what is the electric potential at point B? Which the answer was 155 V.
 
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  • #2
meaghan said:
I was thinking that I could maybe set the voltage difference equal to the electric field * charge * distance since the part before asked that if the electric potential at point A is +30.0 V, what is the electric potential at point B? Which the answer was 155 V.
By what logic, theory, or standard equation is a potential equal to field x charge x distance?
 

Related to The Electric Potential Equation

1. What is an electric field?

An electric field is a physical quantity that describes the influence of electric forces on a charged particle. It is a vector field, meaning it has both magnitude and direction, and is created by the presence of electric charges.

2. How do you find the electric field at a point?

The electric field at a point is found by dividing the force exerted on a test charge by the magnitude of the test charge. This can be represented by the equation E = F/q, where E is the electric field, F is the force, and q is the test charge.

3. What units is electric field measured in?

Electric field is typically measured in units of Newtons per Coulomb (N/C) or Volts per meter (V/m). These units represent the force per unit charge and the potential difference per unit distance, respectively.

4. Can the electric field be negative?

Yes, the electric field can have a negative value. This indicates that the direction of the field is opposite to the direction defined as positive. Electric field is a vector quantity, so it can have both positive and negative values depending on the direction of the force.

5. How is electric field related to electric potential?

Electric potential (V) and electric field (E) are related by the equation E = -dV/dr, where dV is the change in electric potential and dr is the change in distance. This means that the electric field is the negative gradient of the electric potential. In other words, the electric field points in the direction of decreasing electric potential.

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