# What is the electric field at the point?

• Ammora
In summary, two equal charges of 7.48 µC are placed at the lower corners of a square with sides of 0.350 m. Using the equation kq/r^2, the magnitude of the electric field at point A, the upper left corner, is calculated as 5.49e+5 N/C. The direction of the field cannot be determined without taking into account the other charge at the lower right corner. It may be possible to use ratios of side lengths instead of trigonometric functions to calculate the force components.
Ammora

## Homework Statement

Two tiny objects with equal charges of 7.48 µC are placed at the two lower corners of a square with sides of 0.350 m, as shown.

Image: http://www.webassign.net/grrphys2/16-p-039.gif

Find the electric field at point A, the upper left corner. (Assume the positive x-direction points to the right.)
magnitude: ____N/C
direction: ____° CCW from the positive x-axis

kq/r^2

## The Attempt at a Solution

I already got the magnitude right, here's what I did:
kq/r^2
= 8.99e9 (7.48e-6)/(.350)^2 = 5.49e+5

Where I'm stuck is the direction.
I tried tan theta = (.350) and that was wrong, and I tried tan theta= (.35/2)/.35 and that was wrong also.
Any help would be greatly appreciated. This is due tonight. Thanks in advance.

Ammora said:
I already got the magnitude right, here's what I did:
kq/r^2
= 8.99e9 (7.48e-6)/(.350)^2 = 5.49e+5

That only accounts for the field due to the charge in the lower left corner. What about the other charge?

Also, because you're dealing with a square with sides parallel to the coordinate axes, you should be able to use ratios of side lengths (similar triangles) in lieu of trig functions to work out the force components, if you wish.

## 1. What is the concept of an electric field?

The electric field is a physical quantity that describes the force experienced by a charged particle when placed in an electric field. It is represented by a vector, with both magnitude and direction.

## 2. How is the electric field at a point calculated?

The electric field at a point is calculated by dividing the force experienced by a test charge placed at that point by the magnitude of the test charge. Mathematically, it can be represented as E = F/q where E is the electric field, F is the force, and q is the test charge.

## 3. What factors affect the strength of an electric field?

The strength of an electric field depends on the magnitude and distance of the source charge, as well as the medium in which the electric field exists. In a vacuum, the electric field decreases with distance from the source charge, but in a medium with a higher permittivity, the electric field may not decrease as rapidly.

## 4. How does the direction of an electric field indicate the direction of the force on a charged particle?

The direction of the electric field at a point is the direction in which a positive test charge would experience a force. Therefore, the direction of the electric field can be used to determine the direction of the force on a charged particle at that point.

## 5. How is the electric field different from the electric potential?

The electric potential is a scalar quantity that describes the potential energy of a charged particle in an electric field. The electric field, on the other hand, is a vector quantity that describes the force experienced by a charged particle in an electric field. In other words, the electric potential is the energy per unit charge, while the electric field is the force per unit charge.

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