NEET Physics Question - Electrostatics

The electric potential at a point (x, y, z) is given by V = – x² y – xz³ + 4 The electric field E at that point is

E = i (2xy + z³) + j x² + k 3xz²

E = i 2xy + j (x² + y²) + k (3xz – y²)

E = i z³ + j xyz + k z²

E = i (2xy – z³) + j xy² + k 3z²x

Solution:

The electric field \(\vec{E}\) is related to the electric potential \(V\) by:

\[
\vec{E} = -\nabla V
\]

Given \(V = -x^2 y - xz^3 + 4\), we calculate the gradient of \(V\):

\[
E_x = -\frac{\partial V}{\partial x} = 2xy + z^3
\]
\[
E_y = -\frac{\partial V}{\partial y} = x^2
\]
\[
E_z = -\frac{\partial V}{\partial z} = 3xz^2
\]

Thus, the electric field is:

\[
\vec{E} = i(2xy + z^3) + j(x^2) + k(3xz^2)
\]

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Solution:

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