# Q&A: What is the strength of the electric field at the position indicated by the dot in (figure 1)?

Below is a list of exercises related to the “What is the strength of the electric field at the position indicated by the dot in (figure 1)?” question. Click on the link to see the detailed solution of the exercise you are interested

## What is the strength of the electric field at the position indicated by the dot in (figure 1)?

Question 1

What is the strength of the electric field at the position indicated by the dot in the figure? What is the direction of the electric field at the position indicated by the dot in the figure? (Assume that x-axis is horizontal and points to the right.)Figure 1 ▼ of 1 +3.0 nC. 5.0 cm 10 cm 10 cm -一3.0 nC

###### Question 2

What is the strength of the electric field at the position indicated by the dot in (Figure 1)?

Question 3

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 that position? Specify the direction as an angle below horizontal.+1.0 nC. 5.0 cm 10 cm 1.0 nC

###### Question 4

What is the strength of the electric field at the position indicated by the dot in (Figure 1) ? Express your answer using two significant figures. What is the direction of the electric field at that position? Specify the direction as an angle below horizontal. Express your answer using two significant figures.

Question 5

Part A Review What is the strength of the electric field at the position indicated by the dot in (Figure 1)? Suppose Q1-7.0 nC and Q2 9.0 nC (Figure 1) Express your answer with the appropriate units You may want to review (Pages 631- 635) Figure E-V alue Units SubmitR Request Answer 5.0 cm Part B 10 cm What is the direction of the electric field at the position indicated by the dot in (Figure 1)? Specify the direction as an angle measured counterclockwise from the positive axis

###### Question 6

Review Part A What are the strength and direction of the electric field at the position indicated by the dot in (Figure 1)? Give your answer in component form. (Assume that x-axis is horizontal and points to the right, and y-axis points upward.) Express your answer in terms of the unit vectors i and j.

Question 7

What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1)?

Specify the strength of the electric field. Let r = 5.4 cm .

E = ?3.0 nC (+ 3.0 nC

###### Question 8

What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1) ?

## What is the strength of the electric field at the position indicated by the dot in (figure 1)?

Part A) What is the strength of the electric field at the position indicated by the

dot in (figure)?

E = ____N/C

Part B) What is the strength of the electric field at the position indicated by the

dot in (figure)? Specify the direction as an angle above the horizontal line.

θ = ?

###### We have two positive particles and we have a dot and we need to figure out the strength of the electric field.

Some dot a positive charge over the positive chart here and we have between two points is five centimet (see figure). So for a positive particle the electric field on the dot here will be pointing outward so the electric field line will be over here so this will be e1 so let’s call this particle one and this will be particle two and then similarly because it’s a positively charged particle it would point out.

We can try to figure out some angles as we solve so this would be. Because this is a symmetric diagram we know that the total over here because the y components of e1 and e2. So this should be e2 the y components of e1 and e2 cancel out because it’s a symmetrical diagram and we’re left with the x component.

To get the distance over here between the two the positive particle and the plot just the neutral particle here or the dot at this location. We need to do a square root so let’s call this d. Let’s say d is the square root of five centimeters so five squared plus five squared so this is just a pathetic pythagoras theorem because this is just a right angle triangle one side so this side is five. This side is also five centimeters and this will give us five square root two centimeters the equation for the electric field is as follows e equals k, k is just the electric field constant, and then q is the charge and then r is the distance between the two the location and the particle so r here actually i should have labeled. This as r so we already found for r q is given in the question so both particles are one. So nano is just one times ten to the negative nine. We know k is 9 times nothing 9 so this is just a constant.

Let’s write out some values that we know so k nine times ten to the nine that is just the constant meter squared over column squared we know q is given. It’s one times ten to the nine coulombs r is what we calculated up here so knowing this we’ll substitute these in so we get E equals 9 times 10 to the 9. The q in this scenario is 1 times 10 negative 9 so this should be negative. We have r 5 square root 2 times 10 to the negative 2 so we’re trying to

convert this to meters squared so on top we have 9 on the bottom we have 5 squared which is 25 square root 2 squared is just 2 and then 10 times 10 to the negative 2 squared is just

1 so we’re left with nine over fifty. Actually this is multiple ten to the negative four. That component will be 50 times 10 to the negative four and this will give us 9 divided by 10, 50 times 10 to the negative four. Otherwise a thousand eight hundred.

Let’s just go back up to the diagram so we can recognize that this triangle over here is one of the special triangles. So it’s one. One of the special triangles is this and we know that the angles here and here are 45 degrees so seeing that this side and this side so this side and this side are the same. You know that the angle over here is 45 degrees, that means the angle over here is 45 degrees and the angle over here is 45 degrees. So this is 45 degrees and this is 45 degrees triangles we can use sine cos but it’s just a shortcut. Then now we know that the y components of e1 e2 cancel each other out because the y component over here is going upward the wacom point is going downward and these e1 e2 are equivalent based on symmetry. We only need to find the x component of e1 and the x component of e2. This direction, we’ll use the coast to evaluate. The e total in our diagram is e1 plus e2 so we know e1, equals e2 so we just have 2 times e1 in the x direction.

So we know that if this will be in the x direction so we’ll do anything about this for a second. So we know we calculated what e is so we have e cos 45 degrees so e is over here what we already calculated based on our formula. Now we’ve taken the x component and multiplied it by two because we have two electric fields. We have two eighteen hundred goes forty five degrees cos 45 degrees let’s go back up here so cos 45 degrees using the special triangle this would just equal adjacent or hypotenuse to adjacent is 1 over square root two.

This will be two times eighteen hundred plus one over square root of two and we end up with 3600 or square root 2 or the electric field and the units are newtons per coulomb.

###### So what is the strength of the electric field at the position indicated by dot.

So we know the strength is 3 600 over square root 2.

We can evaluate this as a decimal point in our calculator if you want to know the direction of the electric field at the position indicated by a dot in the figure and specify the direction as an angle.  So the direction will be in the positive x so the angle is zero degrees. The angle is zero degrees because e t lines up and we know that the y components of e one and e two cancel it out.

So let’s look at the solution so they’ve identified e1 and e2 areequivalent they calculate using the correct formula and they end up with the correct total electric field they provide the incorrect angle so the angle for the separate components is 45 degrees but the angle for the total electric field is zero degrees.

In summary, Correct answer for part A. Incorrect answer for part B. Theta = 0 degrees because the y components of electric fields cancel.

Scroll to Top