Electric force vs electric potential energy

[RaidenXVC]

I'm confused. When I am looking at questions that have me calculate the energy in a point charge the equation uses just the radius (r), but when calculating the force radius squared (r^2) is used. Why is just the radius used in one situation, but radius squared used in another situation?

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[syoung]

Think about the units for PE (energy) and F (force)

Energy (or work) is the force x distance
E = Fd

In this case if you multiply the F by r, it cancels out and leaves you with just r on the bottom for PE

 

[RaidenXVC]

Think about the units for PE (energy) and F (force)

Energy (or work) is the force x distance
E = Fd

In this case if you multiply the F by r, it cancels out and leaves you with just r on the bottom for PE

So this?

 

[NextStepTutor_1]

So, to understand why these two equations look similar, we need to see how potential energy is related to force.

Force = k (q)(q)/ (r^2). Remember that force is a vector.

Based on this, we get the electric field by dividing by the charge:

Electric field = k(q)(q)/(r^2). This is also a vector.

However, to get from electric field to electric potential, the actual derivation requires calculus , which you DONT NEED TO KNOW for the MCAT. Anway we get:

Electric Potential = k(q)/(r). Note, this is a scalar value!

Lastly, electric potential energy is just electric potential * charge, so we get:

Electric potential energy = q*V = k(q)(q) / (r). This is also a scalar.

If you want to know why there is an r on the bottom instead of an r^2 look up the derivation of electric potential from electric field, but know that you DONT NEED TO KNOW IT FOR THE MCAT (the derivation that is). You just need to know how to use the equations.

 

[RaidenXVC]

Thank you

 

[Chrisz]

I'm confused. When I am looking at questions that have me calculate the energy in a point charge the equation uses just the radius (r), but when calculating the force radius squared (r^2) is used. Why is just the radius used in one situation, but radius squared used in another situation?

The overall concept is that we define potential energy at r=infiniti to be zero. The work down by the electric force and change in potential energy of the system must be equal in magnitude but different in signs. Below is mathematical derivation for potential energy from electric force.

Delta PE=-W

dw=FdR=kq1q2/(R^2)dR

PE(at r=infinity)-PE(at R=r)=-W= -integ[r to infinity] kq1q2/(R^2)dR

PE(at R=infinity)-PE(at R=r)=-kq1q2/r

Then we arbitrarily define PE(at r=infinity)=0

so, -PE(at R=r)=-kq1q2/r
PE(at R=r)=kq1q2/r