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Physics Question on electrostatic potential and capacitance

The electric potential at a point (x.y,z)(x. y, z) is given by V=−x2y−xz2+4V = - x^2 y - xz^2 + 4 The electric field at that point is

A

E⃗=i^ 2xy+j^ (x2+y2)+k^(3xz−y2)\vec{E} = \hat{i} \, 2xy + \hat{j} \, (x^2 + y^2) + \hat{k} ( 3xz - y^2)

B

E⃗=i^z3+j^ xyz+k^ z2\vec{E} = \hat{i} z^3 + \hat{j} \,xyz + \hat{k} \,z^2

C

E⃗=i^(2xy−z3)+j^ xy2+k^ 3z2 x\vec{ E} = \hat{i} (2xy - z^3 ) + \hat{j} \,xy^2 + \hat{k} \,3z^2 \, x

D

E⃗=i^(2xy+z3)+j^ x3+k^ 3xz2\vec{ E} = \hat{i} (2xy + z^3 ) + \hat{j} \,x^3 + \hat{k} \,3xz^2

Answer

E⃗=i^(2xy+z3)+j^ x3+k^ 3xz2\vec{ E} = \hat{i} (2xy + z^3 ) + \hat{j} \,x^3 + \hat{k} \,3xz^2

Explanation

Solution

V=−x2y−xz3+4V =- x ^{2} y - xz ^{3}+4 E⃗=−V=−(i^δδx+j^δδy+k^δδz)\vec{ E}=- V =-\left(\hat{ i } \frac{\delta}{\delta x }+\hat{ j } \frac{\delta}{\delta y }+\hat{ k } \frac{\delta}{\delta z }\right) (−x2−xz3+4)\left(- x ^{2}- xz ^{3}+4\right) =(2xy+z3)i^+x2j^+3xz2k^=\left(2 xy + z ^{3}\right) \hat{ i }+ x ^{2} \hat{ j }+3 xz ^{2} \hat{ k }