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# If point charges of equal magnitude are kept at five vertices of a pentagon, how will the resulting electric field at the centre be zero? Please explain it with a mathematical proof.

Asked by Shanthiswaroop Srinivas 2nd May 2018, 6:44 AM
As shown in figure let us assume charges of equal magnitude q Coloumb are placed at each vertex of pentagon.
At Centre O field due to charge at A, FA  = q/ [ (4πε0) r2 ] ; direction of FA is along line AO.

At Centre O field due to charge at B and E are , FB  = q/ [ (4πε0) r2 ] ; FE  = q/ [ (4πε0) r2 ] ;
As shown in figure let us resolve these forces so that one component is along AO and other one perpendicular to AO.
Since magnatiude of FB and FE are same, these perpendicular component cancel each other.

At Centre O field due to charge at C and D are , FC  = q/ [ (4πε0) r2 ] ; FD  = q/ [ (4πε0) r2 ] ;
As shown in figure let us resolve these forces so that one component is along AO and other one perpendicular to AO.
Since magnatiude of FC and FD are same, these perpendicular component cancel each other.

Hence the net field at centre O = FA+FBsin18º+FEsin18º-FCsin54º-FDsin54º
since magnatiude of forces FA,FB,FC,FD,FE are same, we have field at centre O = F(1+2sin18º-2sin54º) = 0
Answered by Expert 2nd May 2018, 12:00 PM
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