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Electric potential due to a point charge = The electrical potential due to point charges so let (x,y) be th e locus of point net potential at point is zero V(x,y,z)= 0 [Q / k.((x-3a)2 + y2 )1/2 ] + [ -2Q / k.((x+3a)2 + y2 )1/2 ] = 0 4*((x-3a)2 + y2 )1/2 - ((x+3a)2 + y2 )1/2 = 0 The locus is a circle : x2 + y2 +9a2 -10.a.x = 0
Electric potential due to a point charge =
The electrical potential due to point charges so let (x,y) be th e locus of point net potential at point is zero V(x,y,z)= 0 [Q / k.((x-3a)2 + y2 )1/2 ] + [ -2Q / k.((x+3a)2 + y2 )1/2 ] = 0 4*((x-3a)2 + y2 )1/2 - ((x+3a)2 + y2 )1/2 = 0 The locus is a circle : x2 + y2 +9a2 -10.a.x = 0
so let (x,y) be th e locus of point net potential at point is zero
V(x,y,z)= 0
[Q / k.((x-3a)2 + y2 )1/2 ] + [ -2Q / k.((x+3a)2 + y2 )1/2 ] = 0
4*((x-3a)2 + y2 )1/2 - ((x+3a)2 + y2 )1/2 = 0
The locus is a circle : x2 + y2 +9a2 -10.a.x = 0
it's got to be an equipotential surface.
consider a unit charge q at a distance x from one of the charges ,now use potential formula KQq/r2 solve
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