Electric Field & Potential: Find Center of Square 2.0 cm

[NicoleKarmo]

Homework Statement

Find the electric field and the potential at the center of a square 2.0 cm with charges of +9.0uC at one corner of the square and with charger of -3.0uC at the remaining 3 corners.

Homework Equations

the sum of Vi= the sum of KQi/ri

The Attempt at a Solution

I saw that the equation was re-arranged this way but I'm not sure how they arrived there:
E= Ea+Eb+Ec+Ed = Ea+Eb+Ec-Eb= Ea+Ec= (K|qa|/r^2 + K|qc|/r^2) towards C

when you plug in the #'s the answer is 5.4x10^8 N/C toward c.
what I don't understand is how they concluded that that was the formula to use.

 

[LowlyPion]

Welcome to PF.

The E-field is a vector field.

The scalar form of the equation can be used to calculate the magnitude of the field at a point. But when you have multiple charges, then you have to pay attention to the direction of all the charges.

When calculating the field at the center then, the opposite corners of the square will act oppositely. Hence when you compare them, the corners with the same charges will offset (equal and like charges equidistant in opposite directions and all that).

That just means then that you need only calculate the magnitude of the different charges because ... well they will make different contributions.

 

[nlightner]

I would first clarify that the problem is referring to electric field and potential due to point charges, rather than a uniform electric field.

Next, I would explain that the formula used is based on the principle of superposition, which states that the total electric field and potential at a point due to multiple point charges is the sum of the individual fields and potentials at that point.

In this case, the electric field and potential at the center of the square can be found by adding the contributions from each individual charge. The formula used is a simplified version of the general formula for electric field, which takes into account the distance between the charges and the point of interest.

I would also mention that the direction of the electric field is toward the negative charge, as opposite charges attract each other.

Lastly, I would provide a step-by-step explanation of how to plug in the given values and arrive at the numerical answer.