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Grade 12Electrostatics

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Profile image of Abhinav Saurabh
12 Years agoGrade 12
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Profile image of siddharth gupta
12 Years ago
20.partially differentiate the potential along x and y axes ,take negative of the expression you obtain.you will be provided with  vectorial expression of electric field.seperate out E vector as magnitude into unit vector.take r=(x+y2)1/2 and deduce aprropriate relation.
Profile image of siddharth gupta
12 Years ago
16.here E is directed along x,y and z axes and is equal to 2i^+2j^+2k^. displacement vector fo(1,1,1) is i^+j^+k^. take dot product of E and d with negative sign.you will obtain a p.d of 6 V between origin and given point i.e answer is 4V.
Profile image of Abhinav Saurabh
12 Years ago
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Profile image of siddharth gupta
12 Years ago
mathematically ∂V/∂x=-Ky and ∂V/∂y=-Kx i.e E=-(-K(yi+xj))=K(yi+xj)​
magnitude of E=K(x2+y2)1/2=Kr.
it is same as explanation provided above and is only its mathematical part.