- #1

pbj_sweg

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## Homework Statement

(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##

## Homework Equations

$$\vec{E} = \frac{Kq}{r^2}$$

$$\text{Pythagorean's Theorem:}~a^2+b^2=c^2$$

$$K = 9\times10^{-9}$$

## 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!