What Is the Electrical Field Magnitude at Each Point Due to the Charges?

[baird.lindsay]

Homework Statement

Two point charges are placed at the opposite corners of a rectangle as shown. What is the Electrical Field magnitude at each point due to the charges?

Homework Equations

Pythagorean theorem, E=kq/r^2

The Attempt at a Solution

r= square root of .8^2 +.4^2 =.89
E= 8.99x10^9 X 4X10^-6 divided by (.89^2)
E=8.99x10^9 X 2X 10^-6 divided by (.89^2)
should I not be dividing by .89^2 but rather the radius of an electron?

I am not sure if A and B should be negative or positive or if I am doing this right..?

 

[Simon Bridge]

A and B are points, they are neither negative nor positive.
However, you do need to bear in mind that the electric field is a vector - you only have the equation for the magnitude.

Note, charge 1 is twice the size of charge 2.
For each point, one charge is twice the distance that the other charge is.
It's worth thinking about this - if you double r, then E goes down by... but what if you also double q? ... halve q?

 

[baird.lindsay]

I don't really understand because to find the vector i need to use square root of e in the x direction squared + e in the y direction squared...

I don't get how to find the vector in the x direction ...I have E= kq/r^2 or 8.99X10^9 X 4X10^-6 / (.400^2)

but my professor crossed off .89^2 so I think I did that wrong?

 

[baird.lindsay]

im getting 226499 N/C for Point A and 125639 N/C for Point B...?

 

[Simon Bridge]

What is the field at point A due to the 2μC charge?
Which direction does it point?

What is the field at point A due to the 4μC charge?
Which direction does it point?

What is the total field vector at point A? (use i-j-k notation if you like)

What is the magnitude of the total field at point A?

 

[baird.lindsay]

what is the field at point a due to the 2μc charge?
Which direction does it point?

112375 n/c in the y direction

what is the field at point a due to the 4μc charge?
Which direction does it point?
56187.5 in the x direction

what is the total field vector at point a? (use i-j-k notation if you like)

112375i + 56187.5j
what is the magnitude of the total field at point a?

125639 nc

?

 

[barryj]

The magnitude at A and B are the same.

 

[Simon Bridge]

@barryj: we'll see...

it's no good to have the answer - you need to have the physics.
Please read the PF rules.

@baird.lindsay:
I think we've identified some confusion here ...

112375 n/c in the y direction

... etc... please show your working.
1. electric field at A due to 2 microCoulomb charge:
$E_1= kq_1/r_1^2$
$k=8.99 \times 10^9 \text{N.m}^{-2} \text{C}^{-2}$
$q_1=2\times 10^{-6}\text{C}$
$r_1=4\text{m}$
$E_1=1123.8\text{N/C}$ direction: "upwards" (note: axis not defined in problem)

Want to try that again?

2. the field at point A due to the 4 microCoulomb charge:
$E_2=kq_2/r_2^2$
... what are each of the values?

3. E vector is

4. magnitude of E ...
note:
if $\vec{E}=E_x\hat{\imath}+ E_y\hat{\jmath}$
then $|\vec{E}|=\sqrt{E_x^2 + E_y^2}$

 

[baird.lindsay]

@barryj: we'll see...

it's no good to have the answer - you need to have the physics.
Please read the PF rules.

@baird.lindsay:
I think we've identified some confusion here ... ... etc... please show your working.
1. electric field at A due to 2 microCoulomb charge:
$E_1= kq_1/r_1^2$
$k=8.99 \times 10^9 \text{N.m}^{-2} \text{C}^{-2}$
$q_1=2\times 10^{-6}\text{C}$
$r_1=4\text{m}$
$E_1=1123.8\text{N/C}$ direction: "upwards" (note: axis not defined in problem)

Want to try that again?

2. the field at point A due to the 4 microCoulomb charge:
$E_2=kq_2/r_2^2$
... what are each of the values?

$E_2= kq_1/r_1^2$
$k=8.99 \times 10^9 \text{N.m}^{-2} \text{C}^{-2}$
$q_2=4\times 10^{-6}\text{C}$
$r_1=8\text{m}$
$E_2=561.9\text{N/C}$ direction: "leftwards"

3. E vector is
$\vec{E}=561.9\hat{\imath}+ 1123.8 \hat{\jmath}$

4. magnitude of E ...
note:
if $\vec{E}=561.9\hat{\imath}+ 1123.8 \hat{\jmath}$
then $|\vec{E}|=\sqrt{561.9^2 + 1123.8^2}$

therefore 1256.4 N/C
is this correct?

 

[Simon Bridge]

therefore 1256.4 N/C
is this correct?

You can edit the quote by inserting extra close and open-quote tags.

Note: the i direction is usually to the right horizontally - the direction of the field at A due to the 4mcC charge is to the left ... i.e. the -i direction. But it makes no difference to the magnitude.

I don't check people's arithmetic - but your method is now correct.
Can you see where your thinking was out before?

Now you can repeat for point B and see if Barryj was right ;)

 

[baird.lindsay]

no they are different...i get it now ..thanks for all the help..this was driving me crazy!

 

[Simon Bridge]

Just take things a step at a time - but notice how you didn't really need anyone to tell you when you had it right?
Did you figure out how you were going wrong the first time?

 

[barryj]

I stand corrected. Humble pie tastes terrible.

 

[Simon Bridge]

Don't beat yourself up.
As you advance in science you will find you are wrong far more times than you are right - that's the only way you can be sure of anything. The trick is to be wrong in the right way... so you can learn from it ;)
http://www.ted.com/talks/kathryn_schulz_on_being_wrong.html

It is a very common tactic to get you familiar with how the equations work by giving you problems where things get doubled and halved. This is a case in point.

Note, charge 1 is twice the size of charge 2.
For each point, one charge is twice the distance that the other charge is.
It's worth thinking about this - if you double r, then E goes down by... (how much?) but what if you also double q? ... halve q?

The point of the exercise is to understand the inverse-square law, not just be able to do the math.