I was wonder how would a graph of electrical field strength away from a spherical -ve charge graph would look. Since E= potential gradient I was guessing that since potential increases away from a negative charge , electric field strength would increase away from a negative charge also ? enter image description here


The potential does increase. This means that the gradient of potential plotted against distance, r, from the charge is always positive. But the gradient, $\frac{dV}{dr},$ keeps decreasing in magnitude – just sketch the graph! The field strength in the r direction is given by$$E=\ –\frac{dV}{dr},$$so the field is in the –r direction and decreases in magnitude the further we go from the negative charge.

  • $\begingroup$ Would it look something like what I uploaded in op ? $\endgroup$ – user122343 Nov 22 '18 at 23:37
  • $\begingroup$ Yes, but as there is radial symmetry, it would be clearer to put the centre of the sphere at $r$ = 0 and to have the $r$ axis in the positive direction only (as it represents the distance from the sphere in any direction). We'd also usually have the graph upside down, reflected about the $r$ axis, the negativity of $E$ showing that the field is directed in the –$r$ direction, that is towards the sphere. $\endgroup$ – Philip Wood Nov 23 '18 at 0:16
  • $\begingroup$ This is not meant unkindly, but have you consulted an introductory textbook? $\endgroup$ – Philip Wood Nov 23 '18 at 0:18

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