Electric potential at any point \( V=-\sqrt{23}\,x+4y+5z \), then magnitude of electric field is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 8
The electric field is the negative gradient of the potential, \( \vec{E} = -\nabla V \), so each component is minus the partial derivative of V with respect to that coordinate.
\( E_x = -\frac{\partial V}{\partial x} = \sqrt{23},\quad E_y = -\frac{\partial V}{\partial y} = -4,\quad E_z = -\frac{\partial V}{\partial z} = -5 \)
\( |\vec{E}| = \sqrt{E_x^{2} + E_y^{2} + E_z^{2}} = \sqrt{23 + 16 + 25} = \sqrt{64} = 8 \)
The unit is V m-1 when V is in volt and x, y, z are in metre.