Question

Electric field at an equatorial point is -kp/r^3. But potential at an equatorial point is 0. According to the relation E=-V/length, this will be wrong. How?

Answer 1

@phi @dan815

Answer 2

@campbell_st @radar

Answer 3

Actually, E = - dV/dr, the rate of change of potential.
You can have E even if V is 0 at a point.

Answer 4

At an arbitrary point a distance R from the origin
\[E= - \frac{ dV }{ dR}\] so \[V = \int\limits E \cdot dR + \text{constant}\]

The key thing is that constant because it means that the zero of potential can be chosen at any point.
In this case \[V=0\] when \[R=r\]

Remember that the zero of potential is an arbitrarily chosen point. Often it is the case that the zero is taken at infinity but that is not the case in this example.

Answer 5

Thank you so much! But, the idea's still not very clear.
Anyone else?

Answer 6

@Farcher I don't quite get it after the integration. :\

Answer 7

The potential at a point is the work done in taking a unit positive charge from an arbitrarily chosen zero of potential to that point.

When you evaluate an indefinite integral you should always add a constant.

Remembering that integration is the reverse of integrations lets assume we differentiate x^2 with respect to x. We get 2x.

So integration 2x gets us to x^2?

But hand on if I had differentiated x^2 + 7 I would still have 2x.

So the integral of 2x is x^2 + constant.

Now in this example you can find the value of that constant by putting a known value of V for a given R.

The other way of doing the integral is to make it a definite integral

\[\int\limits_{V_i}^{Vf} dV = V_f - V_i = \int\limits_{r_i}^{r_f} E \cdot dr\]

Where Vf is the potential at the final position rf and Vi is the potential at the initial position ri.

Answer 8

@Farcher Thank you :)