The net electric field at point A is 6.7x10^4Nm.

[chef99]

Homework Statement

Calculate the net electric field at point A.

Homework Equations

Enet(total)

The Attempt at a Solution

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The electric field at point A will due to charge 2 will point towards charge 2 because charge 2 is negative.
The electric field at point A will due to charge 1 will point towards charge 1 because charge 1 is negative.

Enet(total) = Enetx + Enety
Because q1 is vertical, its x-component is 0. Because q2 is horizontal, its y-component is 0. Therefore,

Enet = kq1/r^2 + kq2/r^2

= (9.0x10^9Nm/C)(-6.0x10^-5C) / (3.0m)^2 + (9.0x10^9Nm/C)(-3.0x10^-5 C) / (3.0m)^2
= -60000Nm
= -6.0x10^4Nm
The electric field at point A due to the other charges is -6.0x10^4Nm


I am unsure if this is correct as this is the first time I've done a question when it hasn't been an equilateral triangle, so I don't know if I have the right idea or not. Also, I am unsure if I need to determine the angle of charge A or not. Any help is greatly appreciated.

**I am new to the forum so apologies for any errors in the format of my question.

 

[PumpkinCougar95]

Enet = kq1/r^2 + kq2/r^2

Do you know that Electric field is a vector, and so you have to add it vectorially?

 

[chef99]

Do you know that Electric field is a vector, and so you have to add it vectorially?

I know the field is a vector, but do I still have to do the x- and y-components first if the direction of the two charges fall on the vertical and horizontal axis, as they appear in the diagram? i.e. Would the x-component for q1 not be zero?

 

[PumpkinCougar95]

Q1 will contribute no x component, Q2 will contribute no y component. But You have to Add them like $$ E_{net} ~=~\sqrt{ {E_x}^2+ {E_y}^2 } $$

 

[chef99]

Q1 will contribute no x component, Q2 will contribute no y component. But You have to Add them like $$ E_{net} ~=~\sqrt{ {E_x}^2+ {E_y}^2 } $$

So Enet = √60000Nm^2 + 30000Nm^2

Enet = 67082
Enet = 6.7 x10^4Nm