Gives
Part
Mid point P(+1C)
Point A(+1C) and B(-1C )
Electric force at centre
FAp=r2kq1q2
Put value
FAP=0.529×109×1×1=36×109N
FBP=−0.529×109×1×1=−36×109N
Fnet=36×109−36×109=0N Part (b)
PointA (+1C) point B(-1C) pointC(+1c)
FAP=rAP2kq1q2
Put value
FAP=(31)29×109×1×1=27N
FBP=−(31)29×109×1×1=−27N
FCP=(31)29×109×1×1=27N
Fnet=FAP+FBP+FCP
Fnet=27−27+27=27N
Part(c)
Centre of force
rAP=rBP=rCP=rDP=21
FAP=rA2pKQ1Q2
Put value
FAP=(21)29×109×1×1=18N
FBP=−(21)29×109×1×1=−18N
FCP=(21)29×109×1×1=18N
FDP=−(21)29×109×1×1=−18N
Fnet=FAP+FBP+FCP+FDP Put value
Fnet=0N
Part(d)
(a)
Due to centre for Line
Fnet=0N
Due to centre of triangle
Fnet=27N
Due to square at centre
Fnet =0N
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