# Find the Y component of the electric field

1. Apr 1, 2013

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.

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

This is an electric field due to a continuous distribution of charge,

$$E=\int{\frac{kdq}{r^2}\hat{r}_y}$$
$$dq=\frac{Q}{L}$$
$$r=\sqrt{x^2+y^2}=\sqrt{2.25+y^2}$$
$$\hat{r}_y=sin\theta\hat{j} = \frac{y}{r} = \frac{y}{\sqrt{2.25+y^2}}$$
$$k=8.99*10^9 \frac{Nm^2}{C^2}$$

3. 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

2. Apr 1, 2013

### Sunil Simha

You see, θ is supposed to be negative here ( check out my attachment).So what you found there was simply |sinθ| and because θ is negative, you should have multiplied |sinθ| by -1 to get sinθ.

( At least that what I think is the mistake. If others are reading this and feel that something is wrong in my explanation, please feel free to correct it. It'll help me too. Thanks)

File size:
3.9 KB
Views:
67
3. Apr 1, 2013

### SammyS

Staff Emeritus
The problem appears to be with your unit vector, $\displaystyle \ \hat{r}_y\ .$

It should be something like $\displaystyle \ \hat{r}_y=(\pm)\sin\theta\,\hat{j} = -\frac{y}{r}\hat{j} = -\frac{y}{\sqrt{2.25+y^2}}\hat{j}\,,\$ the (±) depending upon how you define θ .

4. Apr 1, 2013

Thanks Sunil and Sammy.

I had suspected that $\displaystyle \theta$ was my problem, but I didn't understand, and still don't understand, why that is.

As the y-value varies from negative to positive, the $\displaystyle sin\theta$ value will also vary from negative to positive.

How do I know which way to define $\displaystyle \theta$? Shouldn't the integral handle the signs since it is varying between -1 & 5 ?

Thanks again

5. Apr 1, 2013

### Sunil Simha

It is convention that theta is taken counterclockwise with respect to the positive x axis. That is, if you were to rotate the x axis in a counterclockwise direction by an angle whose magnitude is θ, then the angle made is +θ.

Instead, if the x axis were rotated in a clockwise direction by the same magnitude, the angle made would be $-θ$.

Though the integral takes care of the change in sign of the angle when y changes from positive or negative, you had defined the angle when y is positive to be positive (whereas it was actually negative then) which the integral faithfully followed

Enjoy Physics!

6. Apr 1, 2013

Ahh, I see! The angle is changing in a clockwise direction as it follows -1 to 5... That makes perfect sense. Thank you, I never would have realized that on my own.

Mac

7. Apr 1, 2013

### SammyS

Staff Emeritus
The unit vector $\ \hat{r}\$ points away from the charge element dq and toward the location, x=1.5 on the x-axis, so when y is negative, the y-component of $\ \hat{r}\$ is positive. On the other hand, when y is positive, the y-component of $\ \hat{r}\$ is negative.

My comment regarding the sign of θ had to do with the fact that without θ being precisely defined, you can't actually state which sign to use for $\ \pm\sin(\theta)\$ as the y-component of $\ \hat{r}\ .$ However, the sign for the y-component of $\ \hat{r}\$ can definitely be assigned when writing it in terms of y .
$\displaystyle \ \hat{r}= \frac{x}{r}\hat{i}-\frac{y}{r}\hat{j} =\frac{1.5}{\sqrt{2.25+y^2}}\hat{i} -\frac{y}{\sqrt{2.25+y^2}}\hat{j}\$​