- #1
MacLaddy
Gold Member
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Hello,
The following problem was on a prior test of mine, and I would like to see where I went wrong. My answer was a negative value, and my instructor made a note that the negative was the only part that was incorrect. Please help me understand why this shouldn't be a negative.
A negative 45 nC charge is distributed uniformly along the y-axis from y=-1.0 to y=5.0. What is the y-component of the electric field at x=1.5 m on the x-axis.
This is an electric field due to a continuous distribution of charge,
$$E=\int{\frac{kdq}{r^2}\hat{r}_y} $$
$$dq=\frac{Q}{L}$$
$$r=\sqrt{x^2+y^2}=\sqrt{2.25+y^2}$$
$$\hat{r}_y=sin\theta\hat{j} = \frac{y}{r} = \frac{y}{\sqrt{2.25+y^2}}$$
$$ k=8.99*10^9 \frac{Nm^2}{C^2}$$
$$ E=\frac{kQ}{L}\int_{-1}^5 \frac{y}{(2.25+y^2)^{\frac{3}{2}}}dy$$
$$ E=\frac{kQ}{L}\left[\frac{-1}{\sqrt{2.25+y^2}} \right]_{-1}^5$$
$$ E=\frac{(8.99*10^9)*(-45*10^{-9})}{6}* \left[\frac{-1}{\sqrt{2.25+5^2}}-\frac{-1}{\sqrt{2.25+(-1)^2}} \right]$$
$$ E=(-67.425)*(0.363) \approx -24.5 $$
My instructor marked that this value should not be negative, and just by intuition it looks as though it should be positive. If anyone can help me see the error in my ways I would appreciate it.
Thanks,
Mac
The following problem was on a prior test of mine, and I would like to see where I went wrong. My answer was a negative value, and my instructor made a note that the negative was the only part that was incorrect. Please help me understand why this shouldn't be a negative.
Homework Statement
A negative 45 nC charge is distributed uniformly along the y-axis from y=-1.0 to y=5.0. What is the y-component of the electric field at x=1.5 m on the x-axis.
Homework Equations
This is an electric field due to a continuous distribution of charge,
$$E=\int{\frac{kdq}{r^2}\hat{r}_y} $$
$$dq=\frac{Q}{L}$$
$$r=\sqrt{x^2+y^2}=\sqrt{2.25+y^2}$$
$$\hat{r}_y=sin\theta\hat{j} = \frac{y}{r} = \frac{y}{\sqrt{2.25+y^2}}$$
$$ k=8.99*10^9 \frac{Nm^2}{C^2}$$
The Attempt at a Solution
$$ E=\frac{kQ}{L}\int_{-1}^5 \frac{y}{(2.25+y^2)^{\frac{3}{2}}}dy$$
$$ E=\frac{kQ}{L}\left[\frac{-1}{\sqrt{2.25+y^2}} \right]_{-1}^5$$
$$ E=\frac{(8.99*10^9)*(-45*10^{-9})}{6}* \left[\frac{-1}{\sqrt{2.25+5^2}}-\frac{-1}{\sqrt{2.25+(-1)^2}} \right]$$
$$ E=(-67.425)*(0.363) \approx -24.5 $$
My instructor marked that this value should not be negative, and just by intuition it looks as though it should be positive. If anyone can help me see the error in my ways I would appreciate it.
Thanks,
Mac