My understanding from this (https://www.toppr.com/ask/content/concept/gold-leaf-electroscope-209365/) source is that -

To identify the nature of charge: The electroscope is charged by a known body (say positively charged body) and then the body is removed. Next, the body under test is brought in contact with the metal cap. If the leaves diverge further, the body has the same charge(positive) and if the leaves come closer to each other, the body has the opposite charge(negative).

Question - Suppose, we had initially charged the gold leaf electroscope by touching a rod of +10C to the metal knob. Since both leaves are of the same size, they will get equal charge viz. +5C & +5C. Now let's take a charged rod of unknown nature (Say it is negatively charged, -20C). Again, -20C will distribute into the leaves as -10C & -10C. Now the net charge on both the leaves will be (+5C -10C = -5C). Therefore, as the magnitude of charge remains the same as before i.e., 5C the angle of deflection between the leaves should also be the same.
(Assumption - Since all the charges on a body get accumulated at the sharpest point, I’ve assumed that when the rod is touched to the electroscope, all the -20C transfers to the electroscope and distributes between the identical gold leaves.)

So, how do we know that the test rod was negative if the deflection remains unchanged even after touching it?


1 Answer 1


Experiments of this type to check the sign of the charge should have the object under test brought close to the cap of the electroscope but not touch it.
This method then removes the ambiguity mentioned in the question.

The only unambiguous conclusion is as a result of an increase in deflection which shows that the charges are of the same sign.
A decrease in deflection will be produced if the the object under test is of the opposite sign or uncharged.

Going back to the question, if a rod with a charge $q$ touches the cap of an electroscope not all of that charge $q$ will be transferred to the electroscope because the object and the electroscope can be considered to be capacitors and so in the final state the charge $q$ is shared between them.

Using a dc voltage source to initially charge an electroscope and produce a deflection will only work for voltages in the hundreds of volts and above rather than tens of volts and below.
As an example the electroscope, which can be used as a voltmeter and is described here, has a non linear scale calibrated in kilovolts.

  • $\begingroup$ I might as well be the first upvote, though Volta’s brilliant innovation of adding the voltage multiplier condenser would also get an upvote if he were still around! $\endgroup$
    – Ed V
    Jul 4, 2023 at 23:23
  • $\begingroup$ Please correct me if I’m wrong, but this is what I understood. So, now I don’t touch the cap of the electroscope instead test the nature of charge by induction. Then since -20C is kept near the cap, the opposite charge that was already on the electroscope will get collected on the cap i.e., +10C. So the charge on both leaves will reduce and thus they show less deflection. (Hence, we conclude that our test rod was negative) $\endgroup$ Jul 5, 2023 at 8:13
  • 2
    $\begingroup$ @ApogeePoint As I pointed out in my answer a reduction in deflection could be due to an oppositely charged or and uncharged object coming close to the cap of the electroscope. $\endgroup$
    – Farcher
    Jul 5, 2023 at 8:20
  • $\begingroup$ @Farcher – Yes, that’s my fault I should’ve included an uncharged body too. But, I was wondering if my reasoning was correct. $\endgroup$ Jul 5, 2023 at 18:19
  • $\begingroup$ @ApogeePoint Yes, bringing a negatively charged object close to the positively charged electroscope will reduce the deflection. $\endgroup$
    – Farcher
    Jul 5, 2023 at 22:30

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