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Yes...How did the point where E=0 …

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Yes...How did the point where E=0 is 2L can you explain me...

Posted by Khushi....? ??‍? 5 years, 3 months ago

CBSE > Class 12 > Physics

5 answers

Khushi....? ??‍? 5 years, 3 months ago

Samajh gue Ayuah thanks.....??

Ayush Vishwakarma?? 5 years, 3 months ago

[-2q]-----------L------------[8q]----x---------P. Keep a test charge in point p. And the force act by 8q = k8q/x^2 and force acted by (-2q) = k2q/(L+x)^2. Then the Enet=E1-E2. And it is given that E=0 then 0=E1-E2.then E1=E2. K8q/x^2= k2q/(x+l)^2. Then solve don't open bracket. It will be soved by square term. And come -2L. Because dirn is negative to positive . Positive has more magnitude and negative has low magnitude. So positve charge is able to apply the force on that point. Agar abhi bhi ni samajh me aaye toh bolna aur accha se samjha denge.

Khushi....? ??‍? 5 years, 3 months ago

Yes Ayush I got that thanks.....

Khushi....? ??‍? 5 years, 3 months ago

Samajh gye Aakash thanku so much......Or lgta h aap naraj ho gye agar haa to I am so sorry....mujhe bs samajh nii aaya tha isiliye puch liye.....?

Aakash Kumar 5 years, 3 months ago

8q<———L ——>(-2q)<------L------->P Put a charge Q at point P Force on Q due to - 2q= - 2kqQ/L² Force due to 8q= 8kqQ/(2L)²=2kqQ/L² Net force on charge Q at P=a big zero and nothing else than that E at P=0 can be concluded. If you don't understand by this method, go to a physics lab and verify it.

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