Question

The electric potential at a point is given by $ V = \left(-5x + 3y + \sqrt{15z} \right) \, \text{volt}$.
The magnitude of the electric field at that point is

• A. \( \frac{\sqrt{3}}{2} \, \text{Vm}^{-1} \)
• B. \( \frac{4}{\sqrt{2}} \, \text{Vm}^{-1} \)
• C. \( \frac{5}{\sqrt{2}} \, \text{Vm}^{-1} \)
• D. \( 7 \, \text{Vm}^{-1} \)

Hint:

To calculate the electric field, take the gradient of the potential and then compute the magnitude.

Answer 1

The Correct Option is D

The electric field \( \mathbf{E} \) is related to the electric potential \( V \) by: \[ \mathbf{E} = - \nabla V \] Taking the partial derivatives of \( V \) with respect to \( x \), \( y \), and \( z \), we get: \[ E_x = - \frac{\partial V}{\partial x} = 5 \, \text{Vm}^{-1}, \quad E_y = - \frac{\partial V}{\partial y} = -3 \, \text{Vm}^{-1}, \quad E_z = - \frac{\partial V}{\partial z} = -\frac{\sqrt{15}}{2\sqrt{z}} \, \text{Vm}^{-1} \] The magnitude of the electric field is: \[ E = \sqrt{E_x^2 + E_y^2 + E_z^2} \] Substituting the values, we find that the magnitude of the electric field is 7 \( \text{Vm}^{-1} \).