Electric field magnitude from three point charges with Q=80nC, a=3m, b=4m

[roam]

Homework Statement

If Q = 80 nC, a = 3.0 m, and b = 4.0 m in the figure, what is the magnitude of the electric field at point P?

[PLAIN]http://img829.imageshack.us/img829/9703/72660756.gif

Homework Equations

$$\vec{E}=k_e \frac{q}{r^2}\hat{r}$$

The Attempt at a Solution

The magnitude of the electric field at P due to the first charge is:

$$E_1 = (8.9 \times 10^9) \frac{(2 \times 80) \times 10^{-9}}{(4^2+3^2)}=56.96$$

due to the second charge it's:

$$E_2 = (8.9 \times 10^9) \frac{(-80) \times 10^{-9}}{(4^2)} = - 44.5$$

And due to charge 3:

$$E_3 = (8.9 \times 10^9) \frac{(2 \times 80) \times 10^{-9}}{(4^2+3^2)}=56.96$$

The total magnitude of the electric field at P would be: $$56.96-44.5+56.96 = 69.3$$

But the correct answer is 47 N/C. Could anyone please show me what is wrong with my calculations?

 

[N-Gin]

You must treat electric field as a vector! Draw the electric field vectors of each charge and with a little bit of trigonometry it should be no problem.

 

[collinsmark]

$$\vec{E}=k_e \frac{q}{r^2}\hat{r}$$

The Attempt at a Solution

The magnitude of the electric field at P due to the first charge is:

$$E_1 = (8.9 \times 10^9) \frac{(2 \times 80) \times 10^{-9}}{(4^2+3^2)}=56.96$$

due to the second charge it's:

$$E_2 = (8.9 \times 10^9) \frac{(-80) \times 10^{-9}}{(4^2)} = - 44.5$$

And due to charge 3:

$$E_3 = (8.9 \times 10^9) \frac{(2 \times 80) \times 10^{-9}}{(4^2+3^2)}=56.96$$

The total magnitude of the electric field at P would be: $$56.96-44.5+56.96 = 69.3$$

But the correct answer is 47 N/C. Could anyone please show me what is wrong with my calculations?

Hello roam,

You need to treat the individual electric field contributions as vectors. Note the unit vector $\hat r$ in your

$$\vec{E}=k_e \frac{q}{r^2}\hat{r}$$

equation. This unit vector is different for each charge you consider. It has both x and y components.

Express each electric field vector in terms of its x and y components, and then add them together.

 

[roam]

Hi Collinsmark

I tried that, it didn't work:

$$\vec{E_1}=56.96 cos (90) \hat{i}+ 56.96 sin (90) \hat{j}$$

$$= 56.96 \hat{j}$$

$$\vec{E_2}=-44.5 cos (0) \hat{i}-44.5 sin (0) \hat{j})$$

$$= -44.5 \hat{i}$$

$$\vec{E_3}=56.96 cos (90) \hat{i}+ 56.96 sin (90) \hat{j}$$

$$= 56.96 \hat{j}$$

And now adding up to find the components of the net electric field vector::

$$E_x= -44.5$$

$$E_y= 56.96+56.96=113.92$$

The resultant would be 122.30 which is wrong, the right answer is the single number 47 N/C. So why am I not getting the right answer?

 

[collinsmark]

Hi Collinsmark

I tried that, it didn't work:

$$\vec{E_1}=56.96 cos (90) \hat{i}+ 56.96 sin (90) \hat{j}$$

$$= 56.96 \hat{j}$$

The direction of $\hat r$ is from the charge, to the test point. So the $\hat r$ direction of the top charge is $\hat r = 4/5 \ \hat \imath - 3/5 \ \hat \jmath$. I'll let you do the rest.

 

[roam]

So $$\hat{r}$$ is a unit vector directed from the charge toward P. But I'm a little bit confused... where did you get "4/5" and "-3/5" from? I know that 5 is the hypotenuse (line from the top charge to P), but why did you divide them like that?

 

[collinsmark]

So $$\hat{r}$$ is a unit vector directed from the charge toward P. But I'm a little bit confused... where did you get "4/5" and "-3/5" from? I know that 5 is the hypotenuse (line from the top charge to P), but why did you divide them like that?

Well, putting it somewhat simply, 'sine' is the opposite over hypotenuse. 'Cosine' is the adjacent over hypotenuse. There is a negative sign associated with the $\hat \jmath$ since that component is pointing down.

$$\sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$$

$$\cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$$