# Electric Field Problem (why don't I multiply by sin and cos?)

• blackbrawler
In summary, the three fixed particles in the figure have charges q1 = q2 = +e and q3 = 2e, with a distance of a = 6.00 µm between them. We are asked to find the magnitude of the net electric field at point P. Using the equation E = k*q/r^2, we calculate the force between q3 and point P (F = 9*10^9 * (3.2/(6^2/2)) * 10^-13). However, we must also calculate the actual distance between q3 and P, which is r = sqrt(a^2/2). This results in a final answer of 160N, rather than the incorrect answer
blackbrawler

## Homework Statement

In the figure the three particles are fixed in place and have charges q1 = q2 = +e and q3 = 2e. Distance a = 6.00 µm. What is the magnitude of the net electric field at point P due to the particles?

E = k*q/r^2

## The Attempt at a Solution

I looked in the back of the book and it shows that I shouldn't be multiplying by (cos45 + sin45) but I don't understand why since Electric Field is a vector value

here is my work

q1 and q2 cancel out so i only calculate q3

F = 9*10^9 * (3.2/(6^2/2)) * 10^-13 * (cos45 + sin45)

simplifies to
F = 160 * sqrt 2 = 226.274N

The actual answer is just 160N so that means I shouldn't multiply by (cos45 + sin45) even though that is the angle.

Edit: picture should be viewable

Last edited:
The image isn't displaying.

Maybe_Memorie said:
The image isn't displaying.

can you see it now?

It's important to know that r is the physical distance between the charge and the point where you want to calculate the field. You can't plug in vector components for r because it's a scaler. I think you assumed that r=6 um, but you actually don't have the value for r yet. You need to find the actual distance between q3 and P.

flatmaster said:
It's important to know that r is the physical distance between the charge and the point where you want to calculate the field. You can't plug in vector components for r because it's a scaler. I think you assumed that r=6 um, but you actually don't have the value for r yet. You need to find the actual distance between q3 and P.

i kind of skipped some of the steps but I did calculate r as sqrt(a^2/2) to get r^2 = 6um^2/2

blackbrawler said:
i kind of skipped some of the steps but I did calculate r as sqrt(a^2/2) to get r^2 = 6um^2/2

Double check that.

flatmaster said:
Double check that.

a^2 = 2r^2

(a^2)/2 = r^2

sqrt((a^2)/2) = r

Electric field is a vector, but the question indicates that it's only asking for the magnitude, so that's why you can disregard the direction.

jackarms said:
Electric field is a vector, but the question indicates that it's only asking for the magnitude, so that's why you can disregard the direction.

but isn't the magnitude effected by the direction or angle?

If I'm applying a force at a 45 degree angle the amount it moves is different than if I apply it at a 90 degree angle.

Depends. For torque it matters because the equation involves a vector product, which necessarily invokes direction. For electric field magnitude only depends on charges and distance, which are both scalars. Angle affects the amount of magnitude in the x and y directions, but not the total magnitude.

blackbrawler said:
F = 9*10^9 * (3.2/(6^2/2)) * 10^-13 * (cos45 + sin45)
There is no logical basis for the term (cos45 + sin45). If the field is E (vector) it has X and Y components |E|cos45 and |E|sin45, but you cannot add those together. To combine perpendicular components of a vector to find its magnitude you use the root-sum-square formula, which, of course, produce the answer |E|.

## 1. Why do we need to use vector components in electric field problems?

The electric field is a vector quantity, meaning it has both magnitude and direction. In order to accurately describe and calculate the electric field at a point, we must use vector components, such as sin and cos, to represent its direction.

## 2. Why can't we simply use scalar values in electric field problems?

Scalar values only have magnitude and do not account for direction. This would not accurately represent the electric field, which is a vector quantity. Using vector components allows us to consider both magnitude and direction.

## 3. How do we determine the direction of the electric field at a point?

The direction of the electric field at a point is determined by the direction of the force that would be exerted on a positive test charge placed at that point. This can be determined using vector components, such as sin and cos, to represent the direction.

## 4. When do we use sin and cos in electric field problems?

Sin and cos are used when calculating the horizontal and vertical components of the electric field. These components are then combined to determine the overall direction and magnitude of the electric field at a point.

## 5. What is the significance of using vector components in electric field problems?

Using vector components allows us to accurately represent and calculate the electric field, which is a vector quantity. It also allows us to determine the direction of the electric field at a point, which is crucial in understanding the behavior and interactions of charged particles.

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