Calculating the magnitude of the total electric field at P

In summary, the electric field at point P due to the first charge has a magnitude of 8 N/C, and points directly to the right. The electric field at point P due to the second charge has a magnitude of 3 N/C, and points directly up.
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ddobre
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Homework Statement


Two electric charges each produce electric fields. At a certain point in space P, the electric field due to the first charge has a magnitude of 8 N/C, and points directly to the right. The electric field at that point due to the second charge has a magnitude of 3 N/C, and points directly up.

(a) What is the magnitude of the total electric field at point P

(b) If an object with a charge of 0.83 coulombs is placed at point P, what is the magnitude of the force it will experience?

Homework Equations


E = k(Q/(r^2))
E = F/q

The Attempt at a Solution


I tried drawing the diagram to get a better understanding of the direction of the fields. I know that I should probably be breaking the vectors into components, but I am having issues due to the lack of information provided in the problem. But, if the charges point directly right towards the point P, and directly up towards P, could I not add the magnitudes of the electric field? I attempted this, finding the answer to be incorrect. I also thought perhaps since I don't have the radius I could try to relate equations to solve for it and other unknown variables, but I found that I could not relate the terms.
 
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Okay
You can't just add electric fields because they are vectors.You need to know the angle between them.

What is the angle between the electric fields at point P
 
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ddobre said:
I tried drawing the diagram to get a better understanding of the direction of the fields. I know that I should probably be breaking the vectors into components, but I am having issues due to the lack of information provided in the problem.
They gave you the magnitudes and directions of the two vectors. Just add them vectorially (place them nose-to-tail)...
 
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  • #4
Right, it was a simple case of vector addition. I have not done that in so long I suppose I forgot it was an option. In any case, the nose-to-tail method worked fine, and the second part was simple enough to find after that. Thank you for the help
 
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1. How do you calculate the total electric field at a specific point P?

The total electric field at a point P is calculated by summing up the contributions from all nearby charges. This can be done using Coulomb's Law, which states that the electric field at a point is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance between the point and the charge.

2. What is the formula for calculating the electric field at a point P?

The formula for calculating the electric field at a point P is E = kQ/r², where E is the electric field, k is the Coulomb's constant (8.99 x 10^9 N m²/C²), Q is the magnitude of the charge, and r is the distance between the charge and the point P.

3. How does the direction of the electric field at a point P relate to the direction of the charge?

The direction of the electric field at a point P is always in the direction of the force that a positive test charge would experience if placed at that point. This means that the direction of the electric field is always away from a positive charge and towards a negative charge.

4. Can the total electric field at a point P be negative?

Yes, the total electric field at a point P can be negative. This can happen if the contributions from the nearby charges are in opposite directions, canceling each other out. In this case, the electric field would be considered negative or in the opposite direction of the resulting field.

5. Are there any other factors that can affect the magnitude of the total electric field at a point P?

Yes, the distance between the charges and the point P, as well as the relative positions and orientations of the charges, can also affect the magnitude of the total electric field. Additionally, the presence of any conductors or other materials in the vicinity can also alter the resulting electric field at point P.

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