The electric potential at a point $(x, y, z)$ is given by $V = -x^2y - xz^3 + 4$. The electric field $\vec{E}$

Question

The electric potential at a point $(x, y, z)$ is given by $V = -x^2y - xz^3 + 4$. The electric field $\vec{E}$ at that point is:

Accepted Answer

The electric field $\vec{E}$ and electric potential $V$ are related by the gradient operation: $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} ight)$ [NCERT Class 12, Sec 2.6.1].

Given the potential $V = -x^2y - xz^3 + 4$ (assuming the trailing '+' implies a constant term like 4 from the standard examination question context):

1.  **Calculate $x$-component ($E_x$):**
    $\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(-x^2y - xz^3 + 4) = -2xy - z^3$
    $E_x = -\frac{\partial V}{\partial x} = -(-2xy - z^3) = 2xy + z^3$

2.  **Calculate $y$-component ($E_y$):**
    $\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(-x^2y - xz^3 + 4) = -x^2$
    $E_y = -\frac{\partial V}{\partial y} = -(-x^2) = x^2$

3.  **Calculate $z$-component ($E_z$):**
    $\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(-x^2y - xz^3 + 4) = -3xz^2$
    $E_z = -\frac{\partial V}{\partial z} = -(-3xz^2) = 3xz^2$

Combining these components, the electric field vector is $\vec{E} = (2xy + z^3)\hat{i} + x^2\hat{j} + 3xz^2\hat{k}$.

Step-by-Step Solution

Practice All PHYSICS Questions