A doubly ionized carbon atom (with charge 2e) is located at the origin of the x

Question

A doubly ionized carbon atom (with charge 2e) is located at the origin of the x axis, and an electron (with charge -e) is placed at x = 2.01 cm. There is one location along the x axis at which the electric field is zero. Give the x coordinate of this point in cm.

____

Assume that the potential is defined to be zero infinitely far away from the particles. Unlike the electric field, the potential will be zero at multiple points near the particles. Find the two points along the x axis at which the potential is zero, and express their locations along the x axis in cm, starting with the point which is farther away from the origin.


Farther Point ____


Closer Point ___

Explanation / Answer

I can't remember the value of e offhand, so I'm going to do this entirely symbolically.

First off, the magnitude of an electric field due to a point charge is E=kq/(r^2)
The net electric field due to 2 (or more) charges is just the sum of the individual fields.

Now, we want to find where the net field will be zero, so
kq(1)/(r(1)^2)+kq(2)/(r(2)^2)=0

Given that this point is along the x-axis, let's think- what's happening at various points?

To the left of the origin, we have the q1 (charge of 2e) generating a positive field (so for a + charge, pushing away) and q2 (charge -e) generating a negative field (for a + charge, pulling towards). This is one possible location.
Between the charges, q1 pushes away from it and q2 pulls towards, so there's no way for them to cancel out.
To the right of q2, we have the same as before, the fields oppose.

I'm relatively certain it's going to be to the right of q2, so let's start there.
If we're looking for a point that's to the right of q2, and we say that point is r away from q1, then the point must be r-d (where d is the distance between the two charges) from q2. This gives us:
kq(1)/(r^2)+kq(2)/((r-d)^2)=0
Which can be simplified to
2ke/(r^2)-ke/((r-d)^2)=0

Solving for r here gives the distance of the point from q1 along the x-axis.

And now, it's all plug and chug.

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