Calculating the Forces on a Charge

[GodlessStar]

Homework Statement

A charge +q is located on the x-axis at a distance h from the xy-plane. There are two uniformly charged bars on the plane as shown in the figure. Calculate the force on the charge q. (Use the results from Ex 2.2)

Relevant Equations

Eqn. 2.2 is located down below.

Here is my attempt. Am I doing it wrong? I am worried I messed up the algebra of it.

 

[PeroK]

It's difficult to read what you've written. Also, it would help if we knew what were the results of exercise 2.2.

 

[Steve4Physics]

Homework Statement: A charge +q is located on the x-axis at a distance h from the xy-plane.

That wouldn't make sense. To match the diagram, the Homework Statement should say "A charge +q is located on the z-axis..."

Relevant Equations: Eqn. 2.2 is located down below.

There is no equation 2.2 shown. There is an expression labelled 2.2 with no explanation of its meaning.

Here is my attempt. Am I doing it wrong? I am worried I messed up the algebra of it. View attachment 358900

Not checked for the same reasons noted by @PeroK in Post #2. But a few comments...

You appear to be using the same symbol for different things, e.g. $\hat M$ (or maybe it's $\hat \mu$) is used for two different unit vectors. You could use subscripts 1 and 2, or L and R.

The total field is $\vec {E_1} + \vec {E_2}$, not $\vec {E_1} - \vec {E_2}$.

Don't forget you are asked for the total force, not the total field.

 

[Gavran]

There are four mistakes.

The first one is the result from Example 2.2. The expression for the electric field a distance $z$ above the midpoint of a bar of length $L$ that carries a uniform line charge density $\lambda$ should be $$ \frac{1}{4\pi\varepsilon_0}\frac{2\lambda L}{z\sqrt{4z^2+L^2}}\hat z $$.

The second one is the length of the right bar in the expression for the electric field produced by the right bar. The length is $2a$.

The third one is the expression for $\vec E_{total}$ in the final equation. It should be $$ \vec E_1+\vec E_2 $$, as it is said in the post #3.

The fourth one is the expression for the unit vector which represents the direction of the electric field produced by the left bar in the final equation. It should be $$ \frac{h\hat k+2a\hat j}{\sqrt{h^2+4a^2}} $$.

At the end the expression for the electric field should be multiplied by the charge $+q$ to get what the force is, as it is said in the post #3.

 

[kuruman]

There are four mistakes.

The first one is the result from Example 2.2. The expression for the electric field a distance $z$ above the midpoint of a bar of length $L$ that carries a uniform line charge density $\lambda$ should be $$ \frac{1}{4\pi\varepsilon_0}\frac{2\lambda L}{z\sqrt{4z^2+L^2}}\hat z $$.

What happened to the y-component of the field? The lines of charge are at different distances from the x-axis.

 

[Gavran]

What happened to the y-component of the field? The lines of charge are at different distances from the x-axis.

The expression is just a hint and the midpoint of the bar is the origin of a three dimensional Cartesian coordinate system where the bar is placed in the xy plane.

The problem solution includes y-component and z-component of the field by using the vectors $$ \frac{h\hat k+2a\hat j}{\sqrt{h^2+4a^2}} $$ and $$ \frac{h\hat k-a\hat j}{\sqrt{h^2+a^2}} $$.

 

[kuruman]

I am worried I messed up the algebra of it.

Then break up the algebra into smaller pieces with fewer symbols. Look at the diagram on the right. You have the superposition of the field from the right $\mathbf E_R$ and from the left $\mathbf E_L.$ For the field from the right, $$\mathbf E_R=E_R\sin(\theta_R)~(-\mathbf {\hat y})+E_R\cos(\theta_R)~(+\mathbf {\hat z})$$ where $$\sin(\theta_R)=\frac{a}{\sqrt{a^2+h^2}}~;~~~\cos(\theta_R)=\frac{h}{\sqrt{a^2+h^2}}.$$You get a similar expression for the field from the left. Add the two vectors together, then replace the trig functions and field magnitudes at the very end.

(Edited to fix mislabeled angles.)

 

[kuruman]

The expression is just a hint and the midpoint of the bar is the origin of a three dimensional Cartesian coordinate system where the bar is placed in the xy plane.

The problem solution includes y-component and z-component of the field by using the vectors $$ \frac{h\hat k+2a\hat j}{\sqrt{h^2+4a^2}} $$ and $$ \frac{h\hat k-a\hat j}{\sqrt{h^2+a^2}} $$.

OK. In your list of OP's mistakes in post #4, when you said that the electric field "should be" it was not clear to me that you were referring to the general case and not specifically to OP's answer.