Electric field question help. Fairly simple but think I'm making small mistakes

[KatherineK]

Homework Statement

The electric field at the point x=5.00 cm and y=0 points in the positive x direction with a magnitude of 10.0 N/C. At the point x=10.0 cm and y=0, the electric field points in the positive x direction with a magnitude of 15.0 N/C. Assuming the field is produced by a single point charge, find (a) it's location and (b) the sign and magnitude of its charge

Homework Equations

E= kq/r^2

The Attempt at a Solution

1.5E1=E2

1.5 kq / x^2 = kq/ (x-5)^2

1.5(x-5)^2 = x^2

used quadratic, got x=27.2 , assumed charge is positive because E field strength increases as you move in the positive x direction.
BUT I think my reasoning might be wrong..

 

[gabbagabbahey]

Hi KatherineK, welcome to PF!

1.5 kq / x^2 = kq/ (x-5)^2

You might be confusing yourself with this notation; what does the "x" in this equation represent?

Are you assuming that the charge is located along the x-axis (y=0)? If so, what is your reasoning to support this assumption?

 

[KatherineK]

I was making x the distance the charge is from the first electric field (10 N /C)
and yes, I am assuming the charge is on the x axis

 

[gabbagabbahey]

I was making x the distance the charge is from the first electric field (10 N /C)

You should probably use a different variable then, so whoever is marking your assignment doesn't think that you are using it to represent the x-coordinate of the point charge.

and yes, I am assuming the charge is on the x axis

Okay, but why?... If you want full credit for your solutions, you should always support any assumptions you make with reasonable arguments.

 

[paranormal]

I would suggest that you double check your calculations and equations to ensure accuracy. It is also important to state the units for the values given in the problem, as the electric field is typically measured in units of N/C. Additionally, it would be helpful to clarify what the variables represent in your equations (e.g. E=electric field, k=Coulomb's constant, q=charge, r=distance).

In terms of your reasoning, it is correct to assume that the charge is positive since the electric field strength increases in the positive x direction. However, it is important to also consider the direction of the electric field at each point. At x=5.00 cm and y=0, the electric field points in the positive x direction, while at x=10.0 cm and y=0, the electric field also points in the positive x direction. This means that the point charge must be located somewhere between these two points, rather than at x=27.2 cm.

In order to find the exact location of the point charge, you can use the equation E=kq/r^2 and substitute in the values given for the electric field and distance at each point. This will give you two equations that you can solve simultaneously to find the value of q and the distance r. Once you have the distance r, you can use it to find the location of the point charge, which will be at a distance of r from the origin in the positive x direction.

Overall, it is important to carefully consider all the given information and use appropriate equations and reasoning to find the correct solution. If you are still unsure, it may be helpful to consult with your teacher or a classmate for further clarification.