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Four particles each having a charge q, are placed on four vertices of a regular pentagon.The distance of each corner from the center is ‘a’ .Find the electric field at the centre of the pentagon.

Aamina , 9 Years ago
Grade 12th pass
anser 3 Answers
krishna priya
If there were 5 charges at each corner then the net electric field at the centre would be 0 since it is a regular polygon .the electric field due to 4 charges will be equal and opposite to the electric field due to the 5th charge so as to make the net electric field at the centre to be 0
Hence electric field at the centre due to 4 charges = electric field at the centre due to the 5th charge
                                                                                                =q/4\pi\epsilon0a2
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Last Activity: 9 Years ago
Ganesh
If there is a charge at 5th vertex of polygon 
   F1+F2+F3+F4+F5=0........eq(1)
From above question there is no charge at 5th vertex so F5 should be eliminated from eq(1)
F1+F2+F3+F4+F5-F5=-F5
therefore F1+F2+F3+F4=-F5
Net force at centre is -ve F5 =kq/a2
 
force at centre of pentagon is equal to -ve of force F5 = 
Last Activity: 7 Years ago
Yash Chourasiya
Hello Student

If there were 5 charges at each corner then the net electric field at the centre would be 0 because it is a regular polygon.
F1 + F2 + F3 + F4 + F5 = 0........eq(1)

F1 + F2 + F3 + F4 = -F5

Electric field at centre due to the 5thcharge =q/4 πϵ0​a2
(Negative sign indicates the direction)

I hope this answer will help you.
Last Activity: 5 Years ago
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