Confused by the fractions in the parenthesis

[patm95]

Homework Statement

Given: see attachment

Five additional electrons are placed at the origin. What is the electric field now.

Homework Equations

Electric field equations of E=q/(4pi*Eo*r^2)

The Attempt at a Solution

I am wondering if the electric field would just be 50% higher because of adding five electrons. However I am somewhat confused by the fractions in the parenthesis. I know that the electric field is going to be a summation of the charges relative to the distance from a point. I am just unsure how to apply that thought in this case due to my misunderstanding of those fractions in the parenthesis. Thanks!

 

[kuruman]

If all you had is five electrons at the origin, what would be the electric field vector as a function of r? If you know the answer, then add that field to the given one and you're done. That's how superposition works.

 

[patm95]

I think I see where u are coming from. So I basically take the 5e/(4pi*ep*r^2) and add it to the original vector?

 

[kuruman]

Yes, add it with unit vector r-hat tacked on it.

 

[paranormal]

The fractions in the parenthesis represent the proportion of the charge (q) to the distance (r) squared. In this case, the equation is used to calculate the electric field at a specific point due to a single charge. Since you are adding five additional electrons at the origin, the total charge at that point will be the sum of the charges of all the electrons. Therefore, the electric field will be higher than before, but it cannot be determined by simply increasing the previous value by 50%. You would need to use the equation to calculate the new electric field at the origin, taking into account the new total charge. If you are still unsure, I recommend seeking further clarification from your instructor or a classmate.