How Do You Calculate Electric Field Strength and Direction?

In summary, the electric field at the position indicated by the dot in Figure 1 is approximately 13289 N/C in the positive direction.
  • #1
miyayeah
27
0

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!
 

Attachments

  • Pr_4.jpg
    Pr_4.jpg
    4.7 KB · Views: 3,408
Physics news on Phys.org
  • #2
miyayeah said:
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?
 
  • #3
NFuller said:
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.
 
  • #4
miyayeah said:
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?
 
  • #5
NFuller said:
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:
xnet= 10788 N/C +1928.8944 N/C
ynet= 3857.7888 N/C

Enet = √(3857.78882N/C+107882N/C) = 13289 N/C.
 
  • #6
You are missing the negative sign on ##q##. Including it gives the field from the negative charge as
$$\mathbf{E_{-}}=-1928.89\text{N/C}\;\hat{x}-3857.79\text{N/C}\;\hat{y}$$
 

Related to How Do You Calculate Electric Field Strength and Direction?

1. What is an electric field?

An electric field is a physical quantity that describes the influence that electric charges have on each other. It is a vector field, meaning it has both magnitude and direction, and is created by the presence and movement of electric charges.

2. How is the electric field at a position calculated?

The electric field at a position is calculated by dividing the force exerted by an electric charge on another charge by the magnitude of the charge. It is also dependent on the distance between the two charges and the medium in which they are located.

3. What is the unit of measurement for electric field?

The unit of measurement for electric field is newtons per coulomb (N/C) in the SI system. However, it can also be measured in volts per meter (V/m) in the CGS system.

4. How does the electric field vary with distance from a source charge?

The electric field is inversely proportional to the square of the distance from a source charge. This means that as the distance increases, the strength of the electric field decreases.

5. What is the difference between electric potential and electric field?

Electric potential is a scalar quantity that describes the potential energy of a charged particle at a specific point in an electric field. It is the potential energy per unit charge. Electric field, on the other hand, is a vector quantity that describes the force per unit charge acting on a charged particle. In other words, electric potential is a measure of the energy stored in an electric field, while electric field is a measure of the force exerted by an electric field.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
365
  • Introductory Physics Homework Help
Replies
5
Views
932
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
859
  • Introductory Physics Homework Help
Replies
2
Views
608
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
274
  • Introductory Physics Homework Help
Replies
10
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
2K
Back
Top