Solve Electric Field for (x,y) Position

[Tina20]

Homework Statement

A -18.3 nC charge is located at position (x,y) = (3.50 cm, 1.75 cm). At what (x,y) position is the electric field -209150.0i? (Enter your x position first followed by the y position.)

Homework Equations

E = (1/4pi epsilon naught)(q/r^2)

Epsilon naught = 8.85x10^-12 C^2/N*m^2

The Attempt at a Solution

I substituted the charge and electric field position and solved for r = 3.1194x10^-12 m

What do I do form there?

 

[collinsmark]

Homework Statement

A -18.3 nC charge is located at position (x,y) = (3.50 cm, 1.75 cm). At what (x,y) position is the electric field -209150.0i? (Enter your x position first followed by the y position.)

What units is the "-209150.0i" electric field in?

Homework Equations

E = (1/4pi epsilon naught)(q/r^2)

Epsilon naught = 8.85x10^-12 C^2/N*m^2

The Attempt at a Solution

I substituted the charge and electric field position and solved for r = 3.1194x10^-12 m

You might want to recheck your math. I got a different value.

What do I do form there?

I assume that the "i" in -209150.0i represents the unit vector along the x-axis. Since this component of the the electric field is negative, this component of the electric field points along the negative x-axis.

So once you find the correct magnitude of r, you'll know how far away the point is from the charge. And since you know where the charge is (and you know that the charge is negative), where in relation to the charge would the electric field point along the negative x-axis?

 

[Tina20]

I checked my math and my r value was wrong. So now that I have an r value of 0.0280m and it is in the negative direction, that means 0.0350m-0.0280m = 0.00695m along the x-axis. Would the y coordinate stay the same?

I used 0.00695i and 0.0175j as the x and y coordinates but that is wrong. Any other hints?

 

[collinsmark]

I checked my math and my r value was wrong. So now that I have an r value of 0.0280m and it is in the negative direction, that means 0.0350m-0.0280m = 0.00695m along the x-axis.

Don't forget that the charge itself is negative.

For a positive charge, the electric field lines radiate away from the [positive] charge. In other words, electric field lines point away from a positive charge. Things are just the opposite for a negative charge. So for a negative charge the direction of the electric field lines are in the direction of ...

So given that, is your point of interest to the left or right of the charge?

Would the y coordinate stay the same?

What do you think? (Okay, here's a hint: yes. The y-coordinate stays the same )

 

[Tina20]

So since the charge is negative, the field points towards the negative charge, so my x position is actually 0.0350m + 0.00695m

 

[Tina20]

It worked, thank you so much for your help! just in time too, deadline is midnight :)