Solve -3C, 8C, 4C Electrostatics Problem

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SUMMARY

The problem involves three small identical balls with charges of -3C, 8C, and 4C. When these balls are brought into contact, the total charge is calculated using the formula q = (q1 + q2 + q3) / 3, resulting in a charge of approximately 3C on each ball after redistribution. This charge redistribution occurs because the balls do not retain memory of their initial charges once they are separated. The discussion clarifies that the initial charges do not return after separation, as the balls become neutralized during contact.

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Homework Statement



three small identical balls have charges -3C,8C and 4C respectively. they are brought in contact and separated. calculate charge on each ball?

Homework Equations



q=(q1+q2+q3)/3

The Attempt at a Solution


what is the need for dividing by 3. why not simply add the charges q1+q2+q3
 
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logearav said:
what is the need for dividing by 3. why not simply add the charges q1+q2+q3
That would give you the total charge, not the charge per ball.
 
sir, it has already been mentioned about the charge possessed by each individual ball. When they come in contact some redistribution of charge occurs. if they are separated won't they get their initial charges?
 
logearav said:
if they are separated won't they get their initial charges?
No, why would they? They retain no "memory" of their initial charges.

Here's an example that might make it clear. Imagine two identical balls, with charges of +5 and -5 units. When brought together the charge redistributes, adding to zero. There's no longer a charge on the balls, they are neutral. If you had two uncharged balls together, would you expect them to gain a charge as they are separated?
 
Thanks sir. i understood now
 

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