# Electric Fields: Magnitude and Direction.

## Homework Statement

Calculate the magnitude and direction of the electric field at point Z in
Figure 13, due to the charged spheres at points X and Y. E = kq1/r^2

## The Attempt at a Solution

Determine each vector component:

Ex = (9.0x10^9 Nm^2/C^2)(50.0 x 10^-6 C) / (0.75m)^2
Ex = 8.0 x 10^5 N/C

Ey = (9.0x10^9 Nm^2/C^2)(10.0 x 10^-6 C) / (0.30m)^2
Ey = 1.0 x 10^6 N/C

Determine sum of vectors:

Ez = ∑E = Ex + Ey = 8.0 x 10^5 N/C + 10 x 10^5 N/C = 1.8 x 10^6 N/C

The answer should be 2.0 x 10^5 N/C

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ehild
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E = kq1/r^2

## The Attempt at a Solution

Determine each vector component:

Ex = (9.0x10^9 Nm^2/C^2)(50.0 x 10^-6 C) / (0.75m)^2
Ex = 8.0 x 10^5 N/C

Ey = (9.0x10^9 Nm^2/C^2)(10.0 x 10^-6 C) / (0.30m)^2
Ey = 1.0 x 10^6 N/C

Determine sum of vectors:

Ez = ∑E = Ex + Ey = 8.0 x 10^5 N/C + 10 x 10^5 N/C = 1.8 x 10^6 N/C

The answer should be 2.0 x 10^5 N/C

What do you mean on "left"? What are the directions of he electric fields due to the individual charges? Note that one of them is positive, the other is negative.

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Zondrina
Homework Helper
It should be noted that the electric field vector is given by:

$\vec E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \hat u$

Where $\hat u$ is a unit vector along an axis extending through the particle.

The field lines go out for positive charges and in for negative ones.

Sorry guys but I still can't find out how to get this answer and this is just what the textbook gives.

Zondrina
Homework Helper
Sorry guys but I still can't find out how to get this answer and this is just what the textbook gives.
Place an arbitrary x-y reference frame at $Z$. Which way do the electric field lines point if you were to place the vectors $\vec E_x$ and $\vec E_y$ at point $Z$?

What unit vectors do these directions correspond to?

If you find those unit vectors, $\hat u_x$ and $\hat u_y$, what happens if you add up the electric field vectors now?

ehild
Homework Helper
When you calculate the electric field using the formula E=kq/r^2, you have to substitute q with its sign. What is the direction of the electric field from the +50μC charge at Z? Draw an arrow. Does it point away from the 50μC or towards it?

E from the q=-10μC charge is negative. What is the direction of the electric field? Does its vector point away or towards q?

ehild

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