# Question 5ef26

May 12, 2016

(3) $- 12 \hat{i} V {m}^{-} 1$

#### Explanation:

Electric field $\vec{E}$ is the gradient of potential $V$ and is written as
$\vec{E} = - \nabla V$

In vector form it can be expressed as
vecE=-(hati(del)/(delx)+hatj(del)/(dely)+hatk( del)/(delz))V# ......(1)

It is given that Electric Potential at any point $\left(x , y , z\right)$ is
$V = 3 {x}^{2}$
As the given function of Potential $V$ is independent of $y \mathmr{and} z$, in the equation (1) derivative with respect to $y \mathmr{and} z$ terms vanish. (1) reduces to
$\vec{E} = - \hat{i} \frac{\partial V}{\partial x}$
Inserting values of $V$, we get
$\vec{E} = - \hat{i} \frac{\partial \left(3 {x}^{2}\right)}{\partial x}$
or $\vec{E} = - 6 x \hat{i}$ .....(2)

To find the electric field at the desired point $\left(2 , 0 , 1\right)$, inserting $x = 2$ in (2) above, (distance in $m$)
$\vec{E} = - \left(6 \times 2\right) \hat{i} = - 12 \hat{i} V {m}^{-} 1$
assuming electric potential function is given in Volt