# Homework Help: Magnitude of electric field at a point

1. Aug 16, 2010

### roam

1. The problem statement, all variables and given/known data

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 [Broken]

2. Relevant equations

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

3. 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?

Last edited by a moderator: May 4, 2017
2. Aug 16, 2010

### 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.

3. Aug 16, 2010

### collinsmark

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.

4. Aug 16, 2010

### 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?

5. Aug 16, 2010

### collinsmark

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.

6. Aug 16, 2010

### 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?

7. Aug 16, 2010

### collinsmark

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.

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

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