Electric Fields: Magnitude and Direction.

[OmniNewton]

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.

Homework Equations

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

​

 

[ehild]

Homework Statement

Homework Equations

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.

ehild​

 

[STEMucator]

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.

 

[OmniNewton]

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

 

[STEMucator]

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]

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