# Calculating E Field Magnitude: Contributions from + and - Charges 5cm Above Axis

• sasuke07
In summary, the collective contribution from both the + and - charges located 5 cm above the horizontal axis is 5.4X10^4.
sasuke07

## Homework Statement

What is the magnitude of the collective contribution to the E field at the origin from both the + and - charges located 5 cm above the horizontal axis?

kQ/d^2

## The Attempt at a Solution

Im just having a problem setting up this problem. Could someone help me on getting started

#### Attachments

• charges.pdf
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hi sasuke07!

(try using the X2 button just above the Reply box )

first, find the magnitude of each contribution,

from that, find the x and y components …

what do you get?

I already figured out the magnitued of the the positive charge 5 cm above the axis and got 5.4X10^4. What do you mean by the x and y components. WOuldn't the negative charge have the same magnitude of the positive charge.

sasuke07 said:
WOuldn't the negative charge have the same magnitude of the positive charge.

yes
What do you mean by the x and y components.

draw an arrow at the origin showing the direction of the field from the positive charge

then find the x and y components of that arrow

(and finally you'll add them to the x and y components of the arrow from the negative charge )

so to figure out the x component wouldn't it be Kq/d^2 where k=9X10^9, q is 30X10^9= charge and d would be .05m^2. Or would i have to use cosine and sine to figure out the x and y components and if so could you show me how to set it up.

no, you must use cos and sin, of the angle the arrow makes

so would it be Kq/r^2sintheta
and Kq/r^2Costheta. Where theta would be 90 degrees?

or would theta be 45

θ is the angle of the line from the charge to the origin

awesome so 45 degrees.

So were the equations correct though?

looks ok

so what is the total field from both the positive and the negative charge?​

i got 7.6X10^4 by doing KQ/r^2sin45 and the answer is the same for cos45. BUt i checked the answer and its supposed to be 7.1X10^4. Any suggestions

sasuke07 said:
i got 7.6X10^4 by doing KQ/r^2sin45 and the answer is the same for cos45. BUt i checked the answer and its supposed to be 7.1X10^4. Any suggestions

i think its 5cm because the length of the x and y components is 5cm.

## 1. How do you calculate the electric field magnitude from charges above an axis?

The electric field magnitude from charges above an axis can be calculated by using the equation E = k*q/(r^2), where k is the Coulomb constant (9x10^9 N*m^2/C^2), q is the charge, and r is the distance from the charge to the point where the field is being measured.

## 2. What is the significance of the distance from the charges in calculating the electric field magnitude?

The distance from the charges is a crucial factor in determining the electric field magnitude. As the distance increases, the strength of the electric field decreases, following an inverse square relationship. This means that the field strength is inversely proportional to the square of the distance from the charges.

## 3. How do you take into account contributions from both positive and negative charges when calculating the electric field magnitude?

To take into account contributions from both positive and negative charges, you must first calculate the electric field magnitude from each charge separately using the equation E = k*q/(r^2). Then, add the individual electric field magnitudes together, taking into account the direction and sign of each field. This will give you the total electric field magnitude at the point of interest.

## 4. Why is it important to consider the direction of the electric field when calculating its magnitude?

The direction of the electric field is essential because it determines the direction of the force that a charge will experience when placed in that field. Knowing the direction of the field can also help in predicting the movement and behavior of charged particles in an electric field.

## 5. Can the electric field magnitude be negative?

Yes, the electric field magnitude can be negative. This usually occurs when the charges are of opposite signs and are relatively close to each other. In this case, the electric field magnitude will be negative for points between the charges and positive for points outside the charges, following the direction of the electric field lines.

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