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2 point charges in air at a distance of 20cm from each other interact with a certain force. the distance from each other where these charges should be placed in oil of relative permutivity 5 to obtain the same force of interaction is?

revathy t.nair , 12 Years ago
Grade 12
anser 4 Answers
Aman Bansal

Last Activity: 12 Years ago

Dear Student,

the force of interaction between two ions, separated by a distance, r in a medium ofrelative permittivity εr is given by

\mbox{force} = \frac {z_1z_2e^2}{4\pi \epsilon _0 \epsilon _r r^2}

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Aman Bansal

Askiitian Expert

Aman Bansal

Last Activity: 12 Years ago

Dear Student,

the force of interaction between two ions, separated by a distance, r in a medium ofrelative permittivity εr is given by

\mbox{force} = \frac {z_1z_2e^2}{4\pi \epsilon _0 \epsilon _r r^2}

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

Askiitian Expert

 

shubham sharda

Last Activity: 12 Years ago

just  use the formula q1*q2/4∏ EoEr r^2

Aditya Garg

Last Activity: 12 Years ago

Thats easy! Use : q1.q2/4piEoEr r^2

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