Calculating Electric Field at Point P(0,0'03,0'04)

[Guillem_dlc]

Homework Statement

We have a sphere of radius $R=6\, \textrm{cm}$ charged with a volume density of charge $\rho =10\, \mu \textrm{C}/\textrm{m}^3$ centered at the origin of coordinates. An infinite plane located in the $z=0$-plane will be charged with a surface charge density $\sigma =0,2\, \mu \textrm{C}/\textrm{m}^2$. What is the modulus of the electric field at its point $(0,3,4)\, \textrm{cm}? Answer:$2,87\cdot 10^4\, \textrmV}/\textrm{m}##

Relevant Equations

$E_\sigma=\dfrac{\sigma}{2\varepsilon_0}$

At point $P(0,0'03,0'04)$ the field caused by the sphere is added to the field caused by the plane.
First, $E_\sigma$
$$E_\sigma=\dfrac{\sigma}{2\varepsilon_0}=\dfrac{0,2\cdot 10^{-6}}{2\varepsilon_0}=11299,44\, \textrm{V}/\textrm{m}$$
Then, $E_0$: Because $r<R$:
$$E_0=\dfrac{\rho}{3\varepsilon_0}r=\dfrac{10^{-6}\cdot 10}{3\varepsilon_0}$$
We see that $r=|\vec{r}|=\sqrt{0,03^2+0,04^2}=0,05\, \textrm{m}$ because $\vec{r}=(0,0'03,0'04)-(0,0,0)=(0,0'03,0'04)$. Then,
$$E_0=\dfrac{10\cdot 10^{-6}}{3\varepsilon_0}\cdot 0,05=18832,39\, \textrm{V}/\textrm{m}$$
Finally,
$$E_T=E_\sigma +E_0=18832,39+11299,44=30131,83\, \textrm{V}/\textrm{m}$$

I don't get the solution I should give, should I have done modulus instead of direct summation?

 

[TSny]

Do the two electric fields $\vec E_0$ and $\vec E_{\sigma}$ point in the same direction at the point (0, 3 cm, 4 cm)?

 

[Guillem_dlc]

Do the two electric fields $\vec E_0$ and $\vec E_{\sigma}$ point in the same direction at the point (0, 3 cm, 4cm)?

No. $\vec E_{\sigma}$ is vertical, and the other has vertical and horizontal components, doesn't it?

 

[TSny]

No. $\vec E_{\sigma}$ is vertical, and the other has vertical and horizontal components, doesn't it?

Right.

 

[PeroK]

No. $\vec E_{\sigma}$ is vertical, and the other has vertical and horizontal components, doesn't it?

I pointed out to you yesterday that the electric field is a vector and you will get things wrong if you don't use vector components:

https://www.physicsforums.com/threa...field-created-by-a-ring.1013418/#post-6612173

 

[Guillem_dlc]

Right.

But how do I separate the $\vec E_0$? The formula is:
$$E_0=\dfrac{\rho}{3\varepsilon_0}r$$

 

[TSny]

But how do I separate the $\vec E_0$?

To find the components of $\vec E_0$ use your knowledge of the direction of $\vec E_0$. A vector diagram can be helpful.

 

[Guillem_dlc]

To find the components of $\vec E_0$ use your knowledge of the direction of $\vec E_0$. A vector diagram can be helpful.

Right. With sine and cosine, right? And I find the angle with the point. Then I multiply the result I have by these and it will give me the two axes, won't it?

 

[TSny]

Right. With sine and cosine, right? And I find the angle with the point. Then I multiply the result I have by these and it will give me the two axes, won't it?

That sounds like the right approach. I would have to see your detailed calculations to be sure.

 

[PeroK]

Right. With sine and cosine, right? And I find the angle with the point. Then I multiply the result I have by these and it will give me the two axes, won't it?

It's simpler to use the unit vector $\hat r = \frac{1}{5}(0, 3, 4)$.