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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 ?

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

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  • $\begingroup$ The potential gradient becomes positive and the electric field is then also negative. $\endgroup$ – Farcher Mar 2 '16 at 7:47
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Look at these two graphs:

enter image description here

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.

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