Obviously on the macroscopic scale, having a test charge being a proton produces no noticeable change in the electric field produced by a charged body.

However, on the microscopic level this would definitely make a difference. So, does a positive test charge (+e) cause the apparent electric field intensity (produced by a change say +10e) to be larger or smaller than the actual value? What would happen if the source charge was negative?

Does it have to do with the attraction or repulsion between unlike and like charges?


"Does it have to do with the attraction or repulsion between unlike and like charges?" Yes.

The electric field always goes out from positive charges and in towards negative charges. The magnitude of the electric field of the +10e charge is independent of the test charge.

\begin{equation} E=\frac{Q}{4\pi\epsilon r^2} \end{equation}

However, the direction of the electric field changes near the vicinity of the test charges. If the test charge is positive then the electric field lines would not meet since the field lines of both the particles are coming out of them and are thus in opposite directions. But if the test charge was negative then the field lines for one particle would be going inside while for the other particle it would be going outside. Since the field lines are going in the same direction (from positive charge towards negative charge) they would meet.

This will happen even if charge placed is $+10e$ while the test charge is only $1e$. Most of the region would be dominated by the field lines of $10e$ however, at some point, close to the $1e$ charge, they will cancel each other out and will not meet- causing the charges to repel.

  • $\begingroup$ So basically, even if the source charge was +2e and the test charge was +e, the electric field intensity itself changes negligibly (due to repulsion between the charges causing change in the field lines) but the position of the source charge would be disturbed? $\endgroup$ – LeroyJD Jul 22 '16 at 7:11
  • $\begingroup$ Also, if the electric field lines of the source charge never reach the test charge (considering like charges), how is it possible to calculate electric field intensity at the point at which the test charge is placed? $\endgroup$ – LeroyJD Jul 22 '16 at 7:14
  • $\begingroup$ of course the placement of another positive charge would cause the particle to move but that is because of the coulomb force. Think of this in the same way as you think of gravity, if you placed a sufficiently heavy body near earth it would cause it to move (in an orbit) but it certainly wouldn't change the $g$ constant of the earth. The electric field is analogous to $g$. $F_G=mg$ and $F_E=qE$ $\endgroup$ – 1D_Particle Jul 22 '16 at 7:20
  • $\begingroup$ Yes you can calculate the electric field intensity at the position of the test charge. If the test charge is placed at a distance of let's say a, then according to the formula above, the magnitude of the electric field would be $E=Q/(4\pi\epsilon a^2)$ $\endgroup$ – 1D_Particle Jul 22 '16 at 7:21
  • $\begingroup$ so suppose the distance between the two particles was a, then at a point of distance "x" from the source charge and distance "a-x" from the test charge, the field would be 0 since the force acting on a charge placed at that point due to the source and test charge would be equal and opposite? $\endgroup$ – LeroyJD Jul 22 '16 at 7:25

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