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enter image description here in above picture,we have a fixed positive charge. this positive charge create electric field in its surrounding.

we consider a special point in left of positive charge.

Is electric field different at that point when we put negative or another positive charge there?

state 1 refer to absence another charge and state 2 refer to presence a negative charge at that point.

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  • $\begingroup$ I think electric field increases and direction of filed is to left.correct? $\endgroup$ – user3728644 Jul 30 '19 at 9:44
  • $\begingroup$ because of this we must use infinitesimal test charge to measure electric field of another charge. because this test charge has no effect on electric field of main charge. correct? $\endgroup$ – user3728644 Jul 30 '19 at 9:47
  • $\begingroup$ Related question Electric field at location of a charge due to the charge itself? $\endgroup$ – Aaron Stevens Jul 30 '19 at 11:47
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If you place a point charge at the point you wish to know the E-field, the E-field is not well defined here. The E-field magnitude becomes arbitrarily large as you approach this point.

If on the other hand you have an extended spherical charged sphere placed at the point you are measuring the E-field, the answer depends. If the free charge distribution in this sphere is spherically symmetric and the sphere is a non-polarizable insulator (its permittivity is $\epsilon_0$), then the value of the E-field is not affected by this sphere. This is because the E-field here is the superposition of the E-field due to the charged sphere and the positive charge on the right, and because of the spherical symmetry, the E-field due to the sphere at the center of this sphere is zero.

In other cases, if you still assume extended charges, but the charge distribution is no longer spherically symmetric or the charged object is made of a dielectric with $\epsilon \ne \epsilon_0$, the E-field will in general be different from the case with only the positive charge.

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  • $\begingroup$ If we put negative sphere at that point, right side of sphere will become more negative than left side. because electric field in conductor must be zero. Correct? $\endgroup$ – user3728644 Jul 30 '19 at 10:03
  • $\begingroup$ you said electric field in center of sphere doesn't change. but electric field in sphere must be zero. $\endgroup$ – user3728644 Jul 30 '19 at 10:06
  • $\begingroup$ why do we use infinitesemal test charge to measure electric field of main charge? $\endgroup$ – user3728644 Jul 30 '19 at 10:09
  • $\begingroup$ The electric field inside has to be zero only if the sphere is a conductor, which is "polarizable." That's why I was careful with the polarizability (or the electric susceptibility) of the charged object in my answer. If the charged object is an insulator, the E-field inside it will not be zero in general. As for the requirement of an infinitesimal test charge, if you don't use a sufficiently small charge, you affect the E-field in its vicinity, so what E-field you are measuring becomes dubious. The E-field blowing near a non-zero point charge is an example of this. $\endgroup$ – Puk Jul 30 '19 at 10:14
  • $\begingroup$ @Puk I think your statement “free charge” on the sphere may be causing confusion. Free charge is normally associated with a conducting surface $\endgroup$ – Bob D Jul 30 '19 at 10:48

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