- #1
exitwound
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Homework Statement
A force of 0.053 N is required to move a charge of 37 mC a distance of 25 cm in an electric field. What is the size of the potential difference between the two points?
Homework Equations
[tex]V_f-V_i=\int \vec{E} \cdot d\vec{S}[/tex]
[tex]E=\frac{F}{q}[/tex]
The Attempt at a Solution
[tex]V_f-V_i=\int \vec{E} \cdot d\vec{S}[/tex]
[tex]\Delta V=Ed[/tex]
[tex]\Delta V=\frac{Fd}{q}[/tex]
[tex]\Delta V=\frac{(.053)(.25)}{37e^-3}= .358 V[/tex]
However, the answer sheet says 360V. Am I missing something stupid? I assume it's a constant electric field and the particle is being moved along the field lines.