Force on a test charge due to two point charges?

[Eclair_de_XII]

Homework Statement

"Let there be two point charges $q_1=3.5μC$ and $q_2=-3.5μC$ located at $(0,0.3 m)$ and $(0,-0.3 m)$ respectively. What force do these two charges exert on a test charge $Q=4.5μC$ at $(0.4 m,0)$?"

*title should read "due to two point charges"* <Moderator's note: title edited>

Homework Equations

The Attempt at a Solution

First off, $∑F_x=0$ because the two point charges cancel each other out in the x-direction. Now I calculate $∑F_y=2F_y$.

$F_y=k\frac{q_1⋅Q}{y^2}=(9.0N_{10^9}⋅m^2⋅C^{-2})(\frac{3.5μC⋅4.5μC}{(0.3m)^2})$
$F_y=(9.0N_{10^{3}}⋅m^2⋅C^{-2})(\frac{15.75C^2}{0.09m^2})=1575000N$
$∑F_y=3200000N$

I am not sure what I'm doing wrong, here.

 

[ehild]

Homework Statement

"Let there be two point charges $q_1=3.5μC$ and $q_2=-3.5μC$ located at $(0,0.3 m)$ and $(0,-0.3 m)$ respectively. What force do these two charges exert on a test charge $Q=4.5μC$ at $(0.4 m,0)$?"

*title should read "due to two point charges"*

Homework Equations

View attachment 209636

The Attempt at a Solution

First off, $∑F_x=0$ because the two point charges cancel each other out in the x-direction. Now I calculate $∑F_y=2F_y$.

$F_y=k\frac{q_1⋅Q}{y^2}=(9.0N_{10^9}⋅m^2⋅C^{-2})(\frac{3.5μC⋅4.5μC}{(0.3m)^2})$
$F_y=(9.0N_{10^{3}}⋅m^2⋅C^{-2})(\frac{15.75C^2}{0.09m^2})=1575000N$
$∑F_y=3200000N$

I am not sure what I'm doing wrong, here.

You used a wrong equation for the y component of the Coulomb force. What is the correct form?
And μC is 10^-6 C, not 10^-3.

 

[Eclair_de_XII]

What is the correct form?

Let me retry this... Is it $F_y=k⋅(\frac{q_1⋅Q}{x^2+y^2})⋅(\frac{y}{\sqrt{x^2+y^2}})$? I want to see if I got the formula right before plugging in any numbers...

 

[ehild]

Let me retry this... Is it $F_y=k⋅(\frac{q_1⋅Q}{x^2+y^2})⋅(\frac{y}{\sqrt{x^2+y^2}})$? I want to see if I got the formula right before plugging in any numbers...

What are x and y?
Your formula is valid in general for the force between a charge at the origin and an other one at point (x,y). Remember, the Coulomb force is inversely proportional to the square of the distance between the point charges. As one coordinate of each charge is zero, your formula might work.

 

[Eclair_de_XII]

What are x and y?

I'm going to set them to $x=\frac{2}{5}m$ and $y=\frac{3}{10}m$.

 

[ehild]

I'm going to set them to $x=\frac{2}{5}m$ and $y=\frac{3}{10}m$.

It will result in correct magnitude.

 

[Eclair_de_XII]

Well, it doesn't matter what sign $x$ is, but I think it would be more accurate if I set $y=-\frac{3}{10}m$ since the force on the test charge is directed at a 270 degree angle.

 

[ehild]

Well, it doesn't matter what sign $x$ is, but I think it would be more accurate if I set $y=-\frac{3}{10}m$ since the force on the test charge is directed at a 270 degree angle.

OK, and how much is it?

 

[Eclair_de_XII]

Let's see...

$F_y=k⋅(y)(\frac{q_1⋅Q}{(x^2+y^2)^\frac{3}{2}})=(9⋅10^9N⋅m^2⋅C^{-2})(-\frac{3}{10}m)\frac{(3.5C)(4.5C)}{(0.5m)^3}⋅(10^{-12})=(9)(10^{-3})(-\frac{3}{10})⋅\frac{15.75}{0.125}=(10^{-3})(-\frac{27}{10})(8)(\frac{63}{4})=-0.34N$

Can anyone confirm if this is correct or not? I kind of jumped the gun and can't enter the answer into my online homework module anymore...

 

[Eclair_de_XII]

Well, thanks for your help @ehild . It really is a better experience learning with other people than with myself. Anyway, even though I screwed myself out of one-eighths of my homework grade, I'm still grateful that I at least started to learn how to do these electrostatic problems more efficiently. Unfortunately, it won't be reflected in my homework grade. But I guess homework is more for learning stuff by practice than for my grade...

 

[ehild]

Let's see...

$F_y=k⋅(y)(\frac{q_1⋅Q}{(x^2+y^2)^\frac{3}{2}})=(9⋅10^9N⋅m^2⋅C^{-2})(-\frac{3}{10}m)\frac{(3.5C)(4.5C)}{(0.5m)^3}⋅(10^{-12})=(9)(10^{-3})(-\frac{3}{10})⋅\frac{15.75}{0.125}=(10^{-3})(-\frac{27}{10})(8)(\frac{63}{4})=-0.34N$

Can anyone confirm if this is correct or not? I kind of jumped the gun and can't enter the answer into my online homework module anymore...

this is the force of one point charge. But you have two of them.

 

[Eclair_de_XII]

So... $∑F_y=-0.68N$?

 

[haruspex]

So... $∑F_y=-0.68N$?

Looks right to me.

 

[Eclair_de_XII]

Thanks.