Question

In an electric field the potential at a point is given by the following relation $V = \dfrac{{343}}{r}$ where $r$ is distance from the origin. The electric field at $r = 3\hat i + 2\hat j + 6\hat k$ is:
(A) $21\hat i + 14\hat j + 42\hat k$
(B) $3\hat i + 2\hat j + 6\hat k$
(C) $\dfrac{1}{7}\left( {3\hat i + 2\hat j + 6\hat k} \right)$
(D) $ - \left( {3\hat i + 2\hat j + 6\hat k} \right)$

Answer 1

Hint: his question uses the concept of electric potential. You need to apply the formula which establishes the relation between the electric potential and electric field. According to the formula, the electric field varies linearly with the potential and inversely with the distance. You are required to substitute the values in the formula and need to calculate the electric field at the given point.

Complete step by step answer:
Given:
The potential at a point is given as $V = \dfrac{{343}}{r}$.
The coordinates of the point is $r = 3\hat i + 2\hat j + 6\hat k$.

As we know that the expression for the electric field is given as shown below,
$\vec E = \dfrac{V}{r} \cdot \hat r$
Here, $V$ is the potential difference and $r$ is the distance from the origin.

Moreover we know that that, $\hat r$can be written as,
$\hat r = \dfrac{{\vec r}}{r}$

So by combining above two expression, we can write it as shown below,
$
\vec E = \dfrac{V}{r} \cdot \dfrac{{\vec r}}{r}\\
\implies \vec E = \dfrac{V}{{{{\left| {\vec r} \right|}^2}}} \cdot \vec r
$

We need to substitute the value of potential in the above expression,
$
\vec E = \dfrac{{\dfrac{{343}}{r}}}{{{{\left| {\vec r} \right|}^2}}} \cdot \vec r\\
\implies \vec E = \dfrac{{343}}{{{{\left| {\vec r} \right|}^3}}} \cdot \vec r.....\left( 1 \right)
$

We can calculate the value of ${\left| {\vec r} \right|^3}$ as shown,
$
\left| {\vec r} \right| = {\left( {\sqrt {{{\left( {3\hat i} \right)}^2} + {{\left( {2\hat j} \right)}^2} + {{\left( {6\hat k} \right)}^2}} } \right)^3}\\
\implies \left| {\vec r} \right| = 343
$

Now we substitute the value of ${\left| {\vec r} \right|^3}$ in equation number (1), to calculate the value of the electric field at the given point.
$
\vec E = \dfrac{{343}}{{343}} \cdot \left( {3\hat i + 2\hat j + 6\hat k} \right)\\
\implies \vec E = 3\hat i + 2\hat j + 6\hat k
$

Therefore, the electric field at point $3\hat i + 2\hat j + 6\hat k$ will be $3\hat i + 2\hat j + 6\hat k$.

Thus, option (B) is correct and electric field is $r = 3\hat i + 2\hat j + 6\hat k$.

Note:
The electric potential is the amount of potential energy at a point per unit charge, and the electric field is the region near a particle that is having some charge. Moreover, in the given question, you can make a mistake in the calculation part, so you need to be careful while working with vector products in the calculation.