# Electric field due to three point charges

• emmanual
In summary, the conversation discusses calculating the intensity of three point charges and determining the resultant intensity using the parallelogram law. The individual electric field vectors are expressed in rectangular coordinates and the challenge lies in adding them up component-wise to find the net electric field. It is suggested to find the separate electric fields first and then add them together for simplicity.
emmanual
Homework Statement
Three point charges q1(+3C), q2(+4C), q3(+3C) are located in a square-shaped form with a distance of 2cm with a 90° cone in every edge(picture in the attached file 39png). Calculate the resultant intensity of point P.
Relevant Equations
Coulomb's law :
F=q1q2/4πεr2
Parallelogram law:
R=√(P²+ Q²+ 2PQcosα)
I've calculated the intensity for every point charge which are
EA = 6.741 x 10¹³ NC¯¹
E
B = 4.494 x 10¹¹ NC¯¹
E
C = 6.741 x 10¹³ NC¯¹
and I am pretty sure about this far but I am struggling to calculate the X-axis intensity and Y-axis intensity to find the entire approximate intensity with the parallelogram law.

#### Attachments

• 2022-02-14 (38).png
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Welcome to PF.

It looks like you have a good start, but those electric field values should be vectors so that you can do the vector sum at P. Can you express all 3 of them as vectors in rectangular coordinates at P so that you can add them up component-wise?

berkeman said:
Welcome to PF.

It looks like you have a good start, but those electric field values should be vectors so that you can do the vector sum at P. Can you express all 3 of them as vectors in rectangular coordinates at P so that you can add them up component-wise?
I've already expressed all 3 in vectors in the first file w the arrows on p and identified them with the EA, EB and EC and which I used to express the resultant intensity too and Idk how I'm supposed to add it up that's where I'm stuck like I know it's supposed to be 2 values, not 3 and IDK how to reduce it into 2 values which will be valid mathematically

Are you saying YDK how to decompose a vector into its components? The magnitude of the vector is represented as the hypotenuse of a right triangle with the x and y components as the two right sides. Have you not seen this before?

SammyS and berkeman
to keep things simple, find separate electrics field of all the charges and in the end add them to get a net electric field due to all the charges, if you want to do other way around its up to you.

## 1. What is an electric field?

An electric field is a physical field that surrounds an electrically charged particle or group of particles. It exerts a force on other charged particles within its vicinity.

## 2. How is the electric field due to three point charges calculated?

The electric field due to three point charges can be calculated using the principle of superposition, which states that the total electric field at a point is equal to the vector sum of the individual electric fields at that point. This can be represented mathematically using Coulomb's law and vector addition.

## 3. What factors affect the strength and direction of the electric field due to three point charges?

The strength and direction of the electric field due to three point charges are affected by the magnitude and sign of the charges, as well as the distance between the charges. The angle between the charges also plays a role in determining the direction of the electric field.

## 4. Can the electric field due to three point charges be negative?

Yes, the electric field due to three point charges can be negative. This occurs when the charges have opposite signs and the resultant electric field is directed towards the negative charge.

## 5. How does the electric field due to three point charges change with distance?

The electric field due to three point charges follows an inverse square law, which means that as the distance from the charges increases, the electric field strength decreases. This is because the electric field spreads out over a larger area as the distance increases.

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