Four charged particles are placed on the vertices of a square. The charges are \(+Q\)(depicted by green) and \(+q\)(depicted by blue). What should be \(\frac{Q}{q}\) so that both the particles having charge \(+Q\) experience net zero electrostatic force?

Please help me with this question. Thanks.

*Swapnil*

## Comments

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TopNewestHave you tried drawing a free body diagram of the +Q charge?

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– Swapnil Das · 1 year ago

Oh yeah. I am stuck somewhere, in the calculations perhaps. Please help.Log in to reply

– Pranshu Gaba · 1 year ago

Kindly share your progress with this problem. What forces did you get on the +Q charge?Log in to reply

– Swapnil Das · 1 year ago

All. I simply applied Vector form of Coulomb's law and plugged values. I did that for both the +Q charges and tried to equate. Doesn't seem to give nice results.Log in to reply

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– Swapnil Das · 1 year ago

As per the book, probably yes. You may check out Fundamentals of Physics.Log in to reply

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@Pranshu Gaba

I think it is correct. What do you thinkBTW, Thanks a lot! – Swapnil Das · 1 year ago

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\[\frac{kQ^2}{2a^2} + \sqrt{2}\frac{k Qq}{a^2} = 0\implies \frac{Q}{q} = \boxed{-2\sqrt{2}}\] – Pranshu Gaba · 1 year ago

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– Rishabh Cool · 1 year ago

Ok..I'll try to upload updated pic maybe later on..Log in to reply

– Swapnil Das · 1 year ago

Of course. Thanks again for help.Log in to reply

– Swapnil Das · 1 year ago

Yeah exactly. Thanks for your time.Log in to reply

– Swapnil Das · 1 year ago

Oh yeah!Log in to reply

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– Swapnil Das · 1 year ago

Oh yeah, you are absolutely correct :)Log in to reply

This can be easily done using physical form of Nash Equilibrium for charges. – Nitesh Chaudhary · 1 year ago

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– Pranshu Gaba · 1 year ago

Wow! I have only encountered Nash Equilibrium in Game theory. I am interested to see how it can be applied in physics. Could you please elaborate on it? Thanks!Log in to reply

– Swapnil Das · 1 year ago

Exactly! We would love to hear from him.Log in to reply

@Nitesh Chaudhary – Swapnil Das · 1 year ago

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