Gives
(a) r1=r2=0.5m
V1=r1kq
Put value
V1=0.59×109×1=18×109V
V2=r2kq
V1=−0.59×109×1=−18×109V Net electric potential at symmet of center
Vnet=V1+V2
Put value
Vnet=18×109−18×109=0V (b)
Centre at Point P
Point AputQ1= +1C
Side(a)=1m
Distance BP=CP=AP=31m
Due to +1C centre of potential
VA=rAPkQ
VA=319×109×1=93V
Similarly point B put charge Q2=-1C
VB=−319×109×1=−93V
Point c put +1C
Vc=319×109×1=93V
Vnet=VA+VB+VC
Vnet=93−93+93=93V (C)
PointA(+1C) point B(-1C)
PointC(+1C) point D(-1C)
Centre point P
AP=BP=CP=DP =21 m
VA=rAPKQ
Put value
VA=219×109×1=92V
Similarly point B
VB=−219×109×1=−92V
Similarly point C
VC=219×109×1=92V
Similarly point D
VD=−219×109×1=−92V
Net potential at centre
V=VA+VB+VC+VD
V=
92V−92V+92V−92V=0V Part(d)
Electric field
EA=r2kq
Put value
EA=.529×109×1=3.6×1010N/C
EB=−.529×109×1=−3.6×1010N/C Enet=EA−EB=0
Electric field of triangle
EA=rAP2kQ
EA=(31)29×109×1=27 N/CEB=−(31)29×109×1=−27N/C
Ec=(31)29×109×1=27N/C
Enet=EA+EB+Ec
Put value
Enet=27N/C
Electric field square
EA=rAP2KQ
EA=(21)29×109×1=18N/C
EB=−(21)29×109×1=−18N/C
EC=(21)29×109×1=18N/C
ED=−(21)29×109×1=−18N/C
Enet=EA+EB+EC+ED
Put value
Enet=0N/c
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