Electrostatics – How to Detect the Nature of Charge Using a Gold Leaf Electroscope

electrostatics

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?

Best Answer

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.

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