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**Hint:**The relation between electric field and potential has a negative sign which implies that the direction of electric field is in the direction of decreasing potential. No work done in moving a test charge over an equipotential surface.

**Formula used:**

We have a relation that is, relation between electric potential and electric field. that is,

$E = - \dfrac{{dV}}{{dr}}$ …........(1)

**Complete step by step answer:**

In a given question there are $x$, $y$, $z$ coordinate axes.

Let we have,$E = {E_x}\hat i + {E_y}\hat j + {E_z}\hat k$ along $x$, $y$ and $z$ axis respectively.

Similarly,$dr = dx\hat i + dy\hat j + dz\hat k$

From equation (1),

$dV = - E.dr$

Electric potential \[\;dV\] is scalar quantity therefore, representing in terms of scalar product or dot product.

Now substitute its values in the above equation we get,

$dV = - ({E_x}\hat i + {E_y}\hat j + {E_z}\hat k).(dx\hat i + dy\hat j + dz\hat k)$

After multiplication using the concept of dot product, that is,

According to this,$\hat i.\hat i = 1$

$\hat j.\hat j = 1$And $\hat k.\hat k = 1$

Also we have, \[\hat i.\hat j = \hat j.\hat k = \hat k.\hat i = 0\]

Then we get, $dV = - {E_x}dx - {E_y}dy - {E_z}dz$

After integration, we can write it as,

This can written as, ${V_{\left( {x,y,z} \right)}} = - {E_x}x - {E_y}y - {E_z}z + c$…………….(2)

Given, the potential is 10v at the origin of coordinates that is at ($0$, $0$, $0$)

Then, at origin, ${V_{\left( {0,0,0} \right)}} = 10V$

At origin equation (2) becomes, ${V_{\left( {0,0,0} \right)}} = - {E_x}x - {E_y}y - {E_z}z + c$

$10 = - {E_x} \times 0 - {E_y} \times 0 - {E_z} \times 0 + c$

$ \Rightarrow c = 10$

Now equation (2) becomes, substituting the value of c,

${V_{\left( {x,y,z} \right)}} = 10 - {E_x}x - {E_y}y - {E_z}z$

Given in the question, 8v at each of the points ($1$,$0$,$0$), ($0$,$1$,$0$) and ($0$,$0$,$1$) that is,

$\Rightarrow {V_{\left( {1,0,0} \right)}} = {V_{(0,1,0)}} = {V_{\left( {0,0,1} \right)}} = 8V$

From these values we can calculate the electric field. We know that, electric field is uniform

Then, ${V_{\left( {1,0,0} \right)}} = 10 - {E_x}x - {E_y}y - {E_z}z = 10 - {E_x} \times 1 - {E_y} \times 0 - {E_z} \times 0$

After simplification, we get $8 = 10 - {E_x}$

$\Rightarrow {E_x} = 2V$

Similarly, ${E_x} = {E_y} = {E_z} = 2V$ since the given electric field is uniform.

Now let us the potential at the point ($1$, $1$, $1$) will be,

$\Rightarrow {V_{\left( {x,y,z} \right)}} = 10 - {E_x}x - {E_y}y - {E_z}z$

On putting the values and we get

$\Rightarrow {V_{\left( {1,1,1} \right)}} = 10 - 2 \times 1 - 2 \times 1 - 2 \times 1$

On simplification we get

$ \Rightarrow {V_{\left( {1,1,1} \right)}} = 10 - 6$

Let us subtract we get

$\therefore {V_{\left( {1,1,1} \right)}} = 4V$

**Thus, the potential at the point ($1$, $1$, $1$) will be $4V$.**

**Note:**Any surface which has the same electric potential at every pint is called equipotential surface.

When a charged particle moves in an electric field, the field exerts a force that can do work on the particle. This work can always be expressed in terms of electrostatic potential energy.

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