Question

The potential function of an electrostatic field is given by $V=2{{x}^{2}}$. Determine the electric field strength at the point ($2\text{ }m,\text{ }0,\text{ }3\text{ }m$).
$\text{A}\text{. }\overrightarrow{E}=4\hat{i}(N{{C}^{-1}})$
$\text{B}\text{. }\overrightarrow{E}=-4\hat{i}(N{{C}^{-1}})$
$\text{C}\text{. }\overrightarrow{E}=8\hat{i}(N{{C}^{-1}})$
$\text{D}\text{. }\,\overrightarrow{E}=-8\hat{i}(N{{C}^{-1}})$

Answer 1

Hint: The electric potential at a point due to a charge distribution is defined as the work to be done to bring a test charge at that point (without acceleration), against the coulombic force, from infinity.
Electric field is the negative gradient of potential. So we can calculate the electric field by calculating the negative gradient of potential.
Formula used:
$\overrightarrow{E}=-\overrightarrow{\nabla }\text{V}$

Complete answer:
We are given
$V=2{{x}^{2}}$.
We know that , negative gradient of potential is electric field.
 $\overrightarrow{E}=-\overrightarrow{\nabla }\text{V}$
$\overrightarrow{E}=-(\dfrac{\partial }{\partial x}\hat{i}+\dfrac{\partial }{\partial y}\hat{j}+\dfrac{\partial }{\partial z}\hat{k})V$
On substituting the value of potential in above formula, we get:
$\overrightarrow{E}=-(\dfrac{\partial }{\partial x}\hat{i}+\dfrac{\partial }{\partial y}\hat{j}+\dfrac{\partial }{\partial z}\hat{k})(2{{x}^{2}})$
$\overrightarrow{E}=-(\dfrac{\partial (2{{x}^{2}})}{\partial x}\hat{i}+\dfrac{\partial (2{{x}^{2}})}{\partial y}\hat{j}+\dfrac{\partial (2{{x}^{2}})}{\partial z}\hat{k})$
$\overrightarrow{E}=-(4x\hat{i}+0\hat{j}+0\hat{k})$
$\overrightarrow{E}=-4x\hat{i}$
To find electric fields at point ($2\text{ }m,\text{ }0,\text{ }3\text{ }m$), we substitute $x=2,y=0$and$z=3$.
$\overrightarrow{E}=-4\times (2)\hat{i}$
$\overrightarrow{E}=-8\hat{i}$

So, the correct answer is “Option D”.

Additional Information:
Electric field Intensity is the force a unit positive charge experiences in that electric field. To determine the electric field at any point, we place a unit positive test charge at that point and determine its experiences. Electric field is a characteristic of the system of charges. It is independent of the charge that we place at any point to determine the electric field.
Electric field is a vector quantity.
According to coulomb’s, electric field due to a single charge in vacuum is
$\vec{E}=\dfrac{1}{4\pi {{\epsilon }_{{}^\circ }}}\dfrac{q}{{{r}^{2}}}\hat{r}$
Where
$\vec{E}=$Electric field
$q=$Charge
$r=$Distance from the point where electric field is to be measured
${{\epsilon }_{{}^\circ }}=$Permittivity of free space

Gradient is a differential operator applied to a vector-valued function to yield a vector whose three components are the partial derivatives of the function with respect to its three variables. The symbol for gradient is $\nabla $.
“The component of electric field in any direction is the negative of the rate of change of the potential in that direction.” For rectangular coordinates, the components of electric field can be written as:
${{E}_{x}}=-\dfrac{\partial V}{\partial x}$, ${{E}_{y}}=-\dfrac{\partial V}{\partial y}$ and ${{E}_{z}}=-\dfrac{\partial V}{\partial z}$.

Note:
A. Direction of the electric field is that in which the potential decreases sharply.
B. Electric field lines are always normal to the equipotential surfaces (Those surfaces where potential is equal).
C. Electric field due to a charge Q does not depend on the test charge q although it is defined in the terms of the test charge.