Question

In a uniform electric field , the potential is \[10V\] at the origin of coordinates and \[8V\] at each of the points\[\left( {1,0,0} \right),\left( {0,1,0} \right)and\left( {0,0,1} \right)\]. The potential at the point \[\left( {1,1,1} \right)\] will be:
\[(A)0\]
\[(B)4V\]
\[(C)8V\]
\[(D)10{{ }}V\]

Answer 1

Hint: Represent the electric field intensity in vector form. Use the relation between electric field intensity and electric potential. The electric potential vector is the dot product of the electric field vector and position vector. Note that the position vector can be found by using the coordinates. Find electric potential for each point given in the problem.

Formula used:
The electric field vector, $\overrightarrow E = {E_x}\widehat i + {E_y}\widehat j + {E_z}\widehat k$
The potential difference, $dV = - \overrightarrow E .\overrightarrow {dr} $
is the position vector.

Complete step by step answer:
Let, the distance between two points is $dr$(small distance) in an uniform electric field $E$, and the electric potential difference between two points is $dV$.
The work done to bring charge from one point to another will be, $( - \overrightarrow {E.} \overrightarrow {dr} )$
According to the definition of the potential difference, this work done is equal to the potential difference between these two points.
Hence, $dV = - \overrightarrow E .\overrightarrow {dr} $ (write in vector form).
The potential vector ,
The electric field vector, $\overrightarrow E = {E_x}\widehat i + {E_y}\widehat j + {E_z}\widehat k$
\[\therefore dV = - ({E_x}\widehat i + {E_y}\widehat j + {E_z}\widehat k).(x\widehat i + y\widehat j + z\widehat k)..........(1)\]
The potential difference at origin is $10V$ and the potential difference each of the points $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ is $8V$
For the point, $(1,0,0)$ the eq. (1) will be,
\[ \Rightarrow (8 - 10) = - ({E_x}\widehat i + {E_y}\widehat j + {E_z}\widehat k).(1\widehat i + 0\widehat j + 0\widehat k)\]
$ \Rightarrow - 2 = - {E_x}\widehat i.\widehat i$
$ \Rightarrow {E_x} = 2V/m$
For the point, $(0,1,0)$ the eq. (1) will be,
\[ \Rightarrow (8 - 10) = - ({E_x}\widehat i + {E_y}\widehat j + {E_z}\widehat k).(0\widehat i + 1\widehat j + 0\widehat k)\]
$ \Rightarrow - 2 = - {E_y}\widehat j.\widehat j$
$ \Rightarrow {E_y} = 2V/m$
For the point, $(0,0,1)$ the eq. (1) will be,
\[ \Rightarrow (8 - 10) = - ({E_x}\widehat i + {E_y}\widehat j + {E_z}\widehat k).(0\widehat i + 0\widehat j + 1\widehat k)\]
$ \Rightarrow - 2 = - {E_z}\widehat k.\widehat k$
$ \Rightarrow {E_z} = 2V/m$
Now for the point $(1,1,1)$, $\overrightarrow E = 2\widehat i + 2\widehat j + 2\widehat k$ and position vector
Now if the potential difference be $V$
So, For the point, $(1,1,1)$ the eq. (1) will be,
\[ \Rightarrow (V - 10) = - (2\widehat i + 2\widehat j + 2\widehat k).(1\widehat i + 1\widehat j + 1\widehat k)\]
$ \Rightarrow V - 10 = - (2 + 2 + 2)$
$ \Rightarrow V = 10 - 6$
$ \Rightarrow V = 4$
So the potential difference at the point $(1,1,1)$ is $4volt$

Hence the right answer is in option $(B)$.

Note: The relation between the potential difference $dV$ and the electric field intensity $E$ is:
$E = - \dfrac{{dV}}{{dr}}$ , $dr$ is the displacement.
This $\dfrac{{dV}}{{dr}}$ is the potential gradient. The negative sign implies that the direction of the electric field intensity along the downwards direction of the electric potential. The unit of the potential gradient is $V.{m^{ - 1}}$ and dimension ${[ML{T^{ - 3}}]^{ - 1}}$ .
Note that, the Electric potential is a scalar quantity but the Electric potential gradient is a vector quantity.