- #1
PFStudent
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Homework Statement
Two large, parallel, conducting plates are 12 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electrostatic force of [tex]3.9 \times {10}^{-15} N[/tex] acts on an electron placed anywhere between the two plates. (Neglect fringing.)
(a) Find the electric field at the position of the electron.
(b) What is the potential difference between the plates?
Homework Equations
[tex]
{\vec{E}}_{p1} = {\frac{\vec{F}_{21}}{q_{2}}}
[/tex]
[tex]
{E_{p1}} = {\frac{|\vec{F}_{21}|}{|q_{2}|}}
[/tex]
[tex]
{E} = {-{\frac{\Delta{V}}{\Delta{r}}}}
[/tex]
The Attempt at a Solution
At first I thought this was a pretty straight-forward problem and got part (a) right but did not understand why part (b) was wrong.
(a)
[tex]
{F_{21}} = 3.9 \times {10}^{-15} N
[/tex]
[tex]
q_{2} = -e
[/tex]
[tex]
{E_{p1}} = {\frac{|\vec{F_{21}}|}{|q_{2}|}}
[/tex]
sig. fig. [itex]\equiv[/itex] 2
[tex]
{E_{p1}} = 2.4 \times {10}^{4} N/C
[/tex]
(b)
[tex]
\Delta{r} = 0.12{\textcolor[rgb]{1.00,1.00,1.00}{.}}m
[/tex]
[tex]
{E} = {-{\frac{\Delta{V}}{\Delta{r}}}}
[/tex]
[tex]
{\Delta{V}} = {-}{\Delta{r}}}}{E}
[/tex]
sig. fig. [itex]\equiv[/itex] 2
[tex]
{\Delta{V}} = -2.9 \times {10}^{3}{\textcolor[rgb]{1.00,1.00,1.00}{.}}V
[/tex]
However, the book answer is: [tex]2.9 \times {10}^{3}{\textcolor[rgb]{1.00,1.00,1.00}{.}}V[/tex].
So, why is the book's answer positive?
Any help is appreciated.
Thanks,
-PFStudent