# Electric field at a point affected by two charges

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1. Feb 10, 2017

### pbj_sweg

1. The problem statement, all variables and given/known data
(a) What is the strength of the electric field at the position indicated by the dot?
(b) What is the direction of the electric field at the position indicated by the dot? Specify direction as an angle measured ccw from the positive x-axis.

$Q_{1} = 4.0 C$
$Q_{2} = 9.0 C$

2. Relevant equations
$$\vec{E} = \frac{Kq}{r^2}$$
$$\text{Pythagorean's Theorem:}~a^2+b^2=c^2$$
$$K = 9\times10^{-9}$$

3. The attempt at a solution
(a) $\vec{E_{1}}$ = effect of $Q_{1}$ on point
$\vec{E_{2}}$ = effect of $Q_{2}$ on point

$\vec{E_{1}} = \frac{KQ_{1}}{r^2} = \frac{K(4~C)}{r^2} = \frac{K(4~C)}{0.05^2} = 1.44\times10^{13} \frac{N}{C}$
$E_{1}\hat{i} = 1.44\times10^{13} \frac{N}{C}$
$E_{1}\hat{j} = 0 ~ \frac{N}{C}$

distance between $Q_{2}~$ and the point was found with Pythagorean's Theorem to be ~0.118 m.
$\vec{E_{2}} = \frac{KQ_{2}}{r^2} = \frac{K(9~C)}{r^2} = \frac{K(9~C)}{0.118^2} = 5.8\times10^{12} \frac{N}{C}$

Breaking $\vec{E_{2}}$ into it's components:
$E_{2}\hat{i} = E_{2}(\sin(\theta)) = 2.596\times10^{12} \frac{N}{C}$
The angle is between the long leg of the triangle and the hypotenuse and was found using tan(0.05/0.1) to get 26.57°.

$E_{2}\hat{j} = E_{2}(\cos(\theta)) = 5.192\times10^{12} \frac{N}{C}$

$E\hat{i} = E_{1}\hat{i} + E_{2}\hat{i} = 1.44\times10^{13} + 2.596\times10^{12} = 1.696\times10^{13}$
$E\hat{j} = E_{1}\hat{j} + E_{2}\hat{j} = 0 + 5.192\times10^{12} = 5.192\times10^{12}$

$| \vec E | = \sqrt{(1.696\times10^{13})^{2}+(5.192\times10^{13})^{2}} = 1.77\times10^{13}$

(b) direction of electric field is $\tan^{-1}\left(\frac{5.192\times10^{12}}{1.696\times10^{13}}\right)$ = 17.01°.

Both parts of the question are incorrect, and I'm 99% sure the second part is incorrect because the components of the electric field on the point are incorrect. Could someone please point out what I've done wrong? Thank you!

2. Feb 10, 2017

### Student100

Is this a typo? Or did you use this value?

3. Feb 10, 2017

### Staff: Mentor

Recalculate your distance value for Q2. You should keep a few extra decimal places in intermediate values such as this distance. This is to prevent rounding/truncation errors from creeping into your significant figures. Only round at the end for your final results.

4. Feb 10, 2017

### pbj_sweg

Sorry, that's a typo. It's to the 9th power. I used the correct value.

Last edited: Feb 10, 2017
5. Feb 10, 2017

### pbj_sweg

Correct distance should be 0.1118 meters instead. I got my final answer to be $1.82\times10^{13}$ which is still wrong. I also didn't round until my final answer. Is there something wrong with my method instead?

6. Feb 10, 2017

### Staff: Mentor

Your result looks good. They may be quibbling about the significant figures though.

Also, what was your angle for the field direction?

Last edited: Feb 10, 2017
7. Feb 10, 2017

### pbj_sweg

I initially got an angle of 17.01° but after recalculating, using the picture you provided, I got 63.25°. Both are wrong . The program usually gives some kind of notification if the issue is with sig figs. Frankly, I think my technique is correct, so maybe MasteringPhysics just has the incorrect answer?

8. Feb 10, 2017

### Staff: Mentor

Your method for calculating the angle was okay. My figure shows the original field contributions by Q1 and Q2, not the net field. I probably should have added that (let me go back and change the figure). The angle should be a bit larger than 17°, but nowhere close to 63°.

9. Feb 10, 2017

### pbj_sweg

I see. Is there anything else wrong with my calculations? Thank you so much for your help!

10. Feb 10, 2017

### Staff: Mentor

I don't see any other problems.

11. Feb 10, 2017

### pbj_sweg

Thank you, seems like the problem is just incorrect.