Section 3.10: Electric Potential and Potential Energy in a Uniform Field
In a uniform electric field, the potential difference between two points is related to the field strength and the displacement along the field lines.
\[ \Delta V = -E \Delta x \]
where \(E\) is the magnitude of the uniform field and \(\Delta x\) is the displacement parallel to the field.
The potential energy of a charge is given by \[ U = qV \]
Example 1
A proton moves through a uniform electric field of \( E = 2000 \, \text{N/C} \). If it moves \( 0.02 \, \text{m} \) parallel to the field, calculate the change in electric potential energy.
Potential difference: \[ \Delta V = -E \Delta x = -(2000)(0.02) = -40 \, \text{V} \]
Change in potential energy: \[ \Delta U = q \Delta V = (1.6 \times 10^{-19})(-40) = -6.4 \times 10^{-18} \, \text{J} \]
The proton’s potential energy decreases; its kinetic energy increases by the same amount.
Practice Problems
- Derive the relationship between electric field and potential difference in a uniform field.
- An electron moves opposite to a uniform field over \(0.01 \, \text{m}\) when \( E = 1000 \, \text{N/C} \). Find the change in potential energy.
- Explain what happens to a charged particle’s kinetic energy when it moves along the field.
- A proton experiences a potential drop of 25 V in a uniform field. Calculate the change in potential energy.
- Explain how energy conservation applies in this situation.