1.

Some equipotential surfaces are shown in figure(29.E3) What can you say about the magnitude and the direction of the electric field?

Answer»

Solution :` (a) The angle between POTENTIAL edl=dv CHANGE in potential =10V=dV `
` As E=rdV `
` (As potential surface) `
` S0, Edl=dv `
` rArr Edlcos(90+30) =-dV `
` rArr E(10xx10^-2)cos120=-dV `
` E=200V//m `
MAKING an angle 120 with yaxis
(b) As ELECTRIC field intensity is rto potential surface
` So, E=r `
` rArr =60vq `
` So, E=v.m=V.M. `


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