Electric fields and potential

E = electric field V = electrostatic potential r = distance ' = derivative

1. E = - (V)'
therefore if E increases V decreases
2. Also if r increases E decreases because E=kq/r^2
3. From 1 & 2, if r increases then V increases
4. But V=kq/r, so if r increases V decreases

Why does the 3rd point contradict with the 4th ?

• Statement 1 should read: $E=-\frac{dV}{dr}$ therefore if $E$ increases $\frac{dV}{dr}$ decreases. – Farcher Mar 2 '16 at 6:57
• @Farcher so if V' decreases it doesn't necessarily mean V decreases ? Could you explain ? – R Kumar Chakravarathy Mar 2 '16 at 7:03
• Explained below as an answer. – Farcher Mar 2 '16 at 8:33

Your fourth statement is incorrect. It should be:

$$V(r) = - \frac{kQ}{r}$$

Note the minus sign, which you omitted. The minus sign means that as $r$ increases $V(r)$ increases because it becomes less negative.

• The potential gradient becomes positive and the electric field is then also negative. – Farcher Mar 2 '16 at 7:47

Look at these two graphs: The potential gradient is negative and gets less negative as $r$ increases.

So minus the potential gradient $=E$ gets less positive.

So it is your first statement which is misleading.