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Question

Point charges of +2μC,+4μC,6μC and +8μC are placed at the corners A,B,C and respectively of a square ABCD of side 0.2 m. Calculate the resultant force on the charge at D.

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Solution



Given four charges placed at corners A, B, C and D of square of side 0.2m.
AD=CD=0.2m
BD=(o.2)2+(0.2)2
=0.283m

So, force on +8μC due to charge at A
F1=14πε02×8×1012(0.2)2
=9×109×4×1010
=3.6N along AD

Force on +8μC due to charge at B
F2=14πε032×8×1012(0.2)2
=9×109×408.16×1012
=3673.47×103
=3.673N along BD

Force on +8μC due to charge at C
F3=9×109×48×1012(0.2)2
=10800×103
=10.8N along AD

Resultant of F1 and F2 is (3.6)2+(10.8)2
F=12.96+116.64
=129.6=11.38N

Now, net force = Resultant of F and F2
=(F)2+(F2)2
=(11.38)2+(3.673)2 along F1
=129.50+1.491
=142.99=11.958N

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