We will now derive the formula for the change in electric potential energy of a charged object q in a constant electric field E when the charged object is moved with displacement Ax. (Please refer to Fig. 1) FAx (Equation 1) F = qE (Equation 2) = -W (Equation 3) W APEetec Substituting Eq. 1 in Eq. 3, APEetec = -FAx (Equation 4) Then subs. Eq. 2 in Eq. 4, APE etec = -qEAx (Eq. 2.1) Which is what we need to derive.

College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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1. What is E in the equation 2.1 derived?
2. Is E in the eq 2.1 always positive? Why?

3. What is Δx in the derived equation?
4. Is it always positive in eq 2.1?

We will now derive the formula for the change in electric potential
energy of a charged object q in a constant electric field E when the
charged object is moved with displacement Ax. (Please refer to Fig. 1)
= FAx (Equation 1)
F = qE (Equation 2)
-W (Equation 3)
W
APEetec
Substituting Eq. 1 in Eq. 3,
APEelec = -FAx (Equation 4)
|
Then subs. Eq. 2 in Eq. 4,
APE elec
- qΕΔx (Εq. 2.1)
Which is what we need to derive.
Transcribed Image Text:We will now derive the formula for the change in electric potential energy of a charged object q in a constant electric field E when the charged object is moved with displacement Ax. (Please refer to Fig. 1) = FAx (Equation 1) F = qE (Equation 2) -W (Equation 3) W APEetec Substituting Eq. 1 in Eq. 3, APEelec = -FAx (Equation 4) | Then subs. Eq. 2 in Eq. 4, APE elec - qΕΔx (Εq. 2.1) Which is what we need to derive.
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