How Do You Calculate Electric Field Strength and Direction?

[miyayeah]

Homework Statement

The diagram is attached. What is the strength of the electric field at the position indicated by the dot in Figure 1? What is the direction of the electric field at the position? Specify the direction as an angle measured clockwise from the positive x axis.

Homework Equations

E=(k⋅q)/r²

The Attempt at a Solution

To solve the first question, I calculated for the individual electric fields from each charge on the point:

E+ = [(8.99⋅10⁹NM²/C²)(3⋅10^-9C)] / (0.05m)²
= 10788 N/C

E- = [(8.99⋅10⁹NM²/C²)(6⋅10^-9C)] / (0.111803m)²]
= 4315 N/C

(0.111803 is from applying trig to the distance between the charges and point.)

Then I found the angle between the line on the y-axis and the line connecting the negative charge to the point:

Θ= tan^-1(0.05/0.1) = 26°
Θ (between imaginary line on x-axis (to the right from negative charge) to the line connecting the negative charge to the point) = 90-26° = 63°

Then I tried to solve the x and y components of the electric field of the negative charge:

(cosΘ)(4315N/C) = 1929 N/C
(sinΘ)(4315N/C) = 3859 N/C

Since there is only x component for the positive charge and to solve for the net electric field at that point, I did:

√((1929 N/C + 10788 N/C)²+(3859N/C)²)

I did not get the answer to the question despite trying out these steps. The correct answer is 9700 N/C. Without knowing the first half of the question I don't think I am able to answer the next half (the direction). Any help would be appreciated!

 

[NFuller]

Then I found the angle between the line on the y-axis and the line connecting the negative charge to the point:

Θ= tan-1(0.05/0.1) = 26°
Θ (between imaginary line on x-axis (to the right from negative charge) to the line connecting the negative charge to the point) = 90-26° = 63°

The problem is that this is not the angle of the electric field vector. The angle is found from the components of the electric field vector, not the distances to the charges. You should first find the x and y components of each vector and add up those components. What do you get when you do this?

 

[miyayeah]

The problem is that this is not the angle of the electric field vector. The angle is found from the components of the electric field vector, not the distances to the charges. You should first find the x and y components of each vector and add up those components. What do you get when you do this?

The electric field vector of the positive charge only has an x component, which I already have calculated. However I am not quite sure how I would find the x and y components of the electric field from the negative charge if I do not have any information about an angle in the first place.

 

[NFuller]

However I am not quite sure how I would find the x and y components of the electric field from the negative charge if I do not have any information about an angle in the first place.

The most direct way is to not worry about finding the angle first and write everything in terms of unit vectors instead. Remember the Electric field vector is given by
$$\mathbf{E}=k\frac{q}{r^{2}}\hat{r}$$
$\hat{r}$ is the unit vector telling you the direction of the electric field. In this case $\hat{r}$ points along the hypotenuse of the triangle created by the two charges and the dot. To find $\hat{r}$ you need to take the components of the vector $\mathbf{r}$ and divide them by the length of $\mathbf{r}$.
$$\hat{r}=\frac{\mathbf{r}}{|\mathbf{r}|}=\frac{5\hat{x}+10\hat{y}}{\sqrt{5^{2}+10^{2}}}=0.447\hat{x}+0.894\hat{y}$$
Now that you have $\hat{r}$ what is the electric field produce by the -6.0nC charge at the location of the dot?

 

[miyayeah]

The most direct way is to not worry about finding the angle first and write everything in terms of unit vectors instead. Remember the Electric field vector is given by
$$\mathbf{E}=k\frac{q}{r^{2}}\hat{r}$$
$\hat{r}$ is the unit vector telling you the direction of the electric field. In this case $\hat{r}$ points along the hypotenuse of the triangle created by the two charges and the dot. To find $\hat{r}$ you need to take the components of the vector $\mathbf{r}$ and divide them by the length of $\mathbf{r}$.
$$\hat{r}=\frac{\mathbf{r}}{|\mathbf{r}|}=\frac{5\hat{x}+10\hat{y}}{\sqrt{5^{2}+10^{2}}}=0.447\hat{x}+0.894\hat{y}$$
Now that you have $\hat{r}$ what is the electric field produce by the -6.0nC charge at the location of the dot?

To find the x component,
E=[(8.99⋅10⁹Nm²/C²)(6⋅10^-9C)(0.447)] / (√0.05²+0.1²m)²
= 1928.8944 N/C

To find the y component,
E=[(8.99⋅10⁹Nm²/C²)(6⋅10^-9C)(0.894)] / (√0.05²+0.1²m)²
= 3857.7888 N/C

However, despite using these values, I still get the same answer as my first attempt:
x_net= 10788 N/C +1928.8944 N/C
y_net= 3857.7888 N/C

E_net = √(3857.7888²N/C+10788²N/C) = 13289 N/C.

 

[NFuller]

You are missing the negative sign on $q$. Including it gives the field from the negative charge as
$$\mathbf{E_{-}}=-1928.89\text{N/C}\;\hat{x}-3857.79\text{N/C}\;\hat{y}$$