# Find the electric field of a square at a given point?

• awilliam_3
In summary, the problem involves two tiny objects with equal charges placed at the two lower corners of a square with sides of 0.580m. The task is to find the electric field at Point A, the upper left corner of the square. The electric field can be calculated using the equation E = K [q1 / d^2], where E is the electric field, K is a constant, q1 is the charge of the object, and d is the distance between the two objects. The solution involves adding the field contributions from each charge, as the field is a vector.
awilliam_3

## Homework Statement

Two tiny objects with equal charges of 6.00uC are placed at the two lower corners of a square with sides of 0.580m.

When facing the square, Point A is the top left corner. Point B is in the top middle. Point C is dead center inside the square.

Find the electric field at Point A, the upper left corner.

E = K [q1 / d^2]

## The Attempt at a Solution

Calculated E1 & E2 using above formula and givens, then took the difference to determine the electric field in N/C.

awilliam_3 said:
Calculated E1 & E2 using above formula and givens, then took the difference to determine the electric field in N/C.
Why did you take the difference? Instead, add the field contribution from each charge, remembering that the field is a vector and must be added as such.

welcome to pf!

hi awilliam_3! welcome to pf!
awilliam_3 said:
Calculated E1 & E2 using above formula and givens, then took the difference to determine the electric field in N/C.

(are the charges the same sign, or opposite sign? if they're opposite, then yes, you subtract)

you can't just add (or subtract) the magnitudes, you have to add (or subtract) the forces as vectors

Doc Al said:
Why did you take the difference? Instead, add the field contribution from each charge, remembering that the field is a vector and must be added as such.

Ah, I see. Thank you.

tiny-tim said:
hi awilliam_3! welcome to pf!

(are the charges the same sign, or opposite sign? if they're opposite, then yes, you subtract)

you can't just add (or subtract) the magnitudes, you have to add (or subtract) the forces as vectors

They are indeed the same sign. Great tip, thank you!

## 1. What is an electric field?

An electric field is a physical field that surrounds an electrically charged object and exerts a force on other charged objects within its vicinity. It is a fundamental concept in the study of electricity and magnetism.

## 2. How is the electric field of a square calculated?

The electric field of a square can be calculated using the formula E = kq/r^2, where k is the Coulomb's constant, q is the charge of the square, and r is the distance from the square to the point where the electric field is being measured.

## 3. What factors affect the strength of the electric field at a given point?

The strength of the electric field at a given point is affected by the magnitude and distribution of the electric charge, as well as the distance between the charged object and the point where the electric field is being measured.

## 4. Why is it important to calculate the electric field of a square at a given point?

Calculating the electric field of a square at a given point can help us understand and predict the behavior of charged objects in the surrounding area. It is also essential in various applications, such as designing circuits and studying the behavior of electromagnetic waves.

## 5. Can the electric field of a square be negative?

Yes, the electric field of a square can be negative. This indicates that the direction of the electric field is opposite to the direction of the force it exerts on other charged objects. In other words, the electric field lines are pointing towards the square, rather than away from it.

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