I'm having some trouble locating the sign error in my line integral calculation for Work done in an electrostatic field. The textbook indicates that my work should be negative, as in $-400~\mu J$, but my answer is $400~ \mu J$. Mostly, i'm confused if I parameterized the direction of the line integral correctly and set the limited correcting in the line integral.
Here's my work:
Given electrostatic field: $\vec{E}=(x/2+2y)~\hat{x} + 2x~\hat{y}~~~(V/m)$
And a test charge $Q = -20 \mu C$ moving in the eletrostatic field from:
(4,2,0) meters to (0,0,0) meters
Here's a plot of the vector field E:
First I find the parametric line equation:
$\vec{r}(t) = (-4t+4)\hat{x} + (2t+2)\hat{y}$
and its derivative is:
$\frac{d\vec{r}(t)}{dt} = -4\hat{x} + 2 \hat{y}$
testing out the line equation I find that:
$\vec{r}(t=0) = 4\hat{x} + 2\hat{y} + 0\hat{z}$
$\vec{r}(t=1) = 0\hat{x} + 0\hat{y} + 0\hat{z}$
which matches the coordinates (4,2,0) to (0,0,0) specified in the problem...
then, i setup the line integral for work, assuming Q is a positive test charge that is moving against the Electric Field and work should be positive:
$W = Q \int^1_0 \vec{E}\Big(\vec{r}(t)\Big) \bullet \frac{d\vec{r}(t)}{dt}~dt$
However, Q is not a positive test charge...but the assumption is that the negative sign of the test charge will correct the orientation as long as I set the problem up for a positive test charge...
substituting $\vec{r}(t)$ into $\vec{E}$:
$\vec{E}\Big(\vec{r}(t)\Big) = \bigg[\frac{-4t+4}{2} + 2(2t+2) \bigg] \hat{x} + \bigg[2(-4t + 4)\bigg]\hat{y}$
$\vec{E}\Big(\vec{r}(t)\Big) = (2t+6)\hat{x} + (-8t+8)\hat{y}$
$\vec{E}\Big(\vec{r}(t)\Big) \bullet \frac{d\vec{r}(t)}{dt} = -24t -8$
finally i can perform the line integral:
$W = (-20\mu) \int \limits_0^1 (-24t -8) dt$
$W = (-20\mu) \bigg[ -12t -8t \bigg]^{1}_{0}$
$W = (-20\mu) (-12 -8)$
$W = (-20\mu)(-20)$
$\boxed{W = 400~ \mu J}$
(Which is the wrong answer... its suppose to be -400 $\mu J$ )
Its -400 so that when I add in the work done by two other paths that are part of a closed loop... it adds to zero:
(0,0,0)->(4,0,0) => $80 \mu J$
(4,0,0)->(4,2,0) => $320 \mu J$
thus: 80 + 320 - 400 = 0.
Your parametric equation is wrong. At $t = 1$, we get \begin{align} \vec{r}(1) &= (-4\cdot1+4)\hat{x} + (2\cdot2+2)\hat{y}\\ &= 0\hat{x} + 4\hat{y} \end{align}
It SHOULD be $$ \vec{r}(t) = (-4t+4)\hat{x} + (-2t+2)\hat{y} $$ with derivative $$ \vec{r'}(t) = -4\hat{x} - 2\hat{y}. $$