Question

Electric potential at any point is $V = - 5x + 3y + \sqrt {15} z$ then the magnitude of the electric field is ______________ $N/C$.
A. $3\sqrt 2 $
B.$4\sqrt 2 $
C. $7$
D. $5\sqrt 2 $

Answer 1

Hint:Electric field at a point is defined as the force that acts on a positive unit charge placed at that point. Electric potential at a point in an electric field is the work required to bring a unit positive charge from infinity to the point. Electric field is a vector and electric potential is a scalar quantity.

Complete step by step answer:
We have seen in vector analysis that a conservative vector field can be derived from a scalar field by taking gradient of the scalar field. Electric field is the negative gradient of electric potential.
$\overrightarrow E = - \overrightarrow \nabla V$
Here the negative sign comes from a physical consideration: the electric field is directed from a region of higher potential to a region of lower potential, whereas the vector $\overrightarrow \nabla V$ is directed in the opposite direction.
$\overrightarrow E = - (\widehat i\dfrac{{\partial V}}{{\partial x}} + \widehat j\dfrac{{\partial V}}{{\partial y}} + \widehat k\dfrac{{\partial V}}{{\partial z}})$

Given, $V = - 5x + 3y + \sqrt {15} z$
$\Rightarrow \overrightarrow E = - ( - \widehat i\dfrac{{\partial (5x)}}{{\partial x}} + \widehat j\dfrac{{\partial (3y)}}{{\partial y}} + \widehat k\dfrac{{\partial (\sqrt {15} )}}{{\partial z}})$
$\Rightarrow \overrightarrow E = + 5\widehat i - 3\widehat j - \sqrt {15} \widehat k$
Magnitude of electric field required
$\left| {\overrightarrow E } \right| = \sqrt {{{(5)}^2} + {{( - 3)}^2} + ( - \sqrt {15} } {)^2}$
$\therefore \left| {\overrightarrow E } \right| = \sqrt {49} = 7$

Hence, the correct answer is option C.

Note:Electric field is the measure of energy per unit area while potential is a property of a field that describes the action of the field upon an object. The electric potential acts as another useful field. It provides an alternative to the electric field in electrostatics. The potential energy for a positive charge increases when it moves against an electric field and decreases when it moves with an electric field; the opposite is true for a negative charge.