Question

If potential (in volts) in a region is expressed as \[~V\left( x\text{ },\text{ }y\text{ },\text{ }z \right)\text{ }=\text{ }6xy\text{ }-\text{ }y+\text{ }2yz,\], the electric field (in N/C) at point \[\left( 1\text{ },\text{ }1\text{ },\text{ }0 \right)\]is –
(A) $-\left( 6\widehat{i}+9\widehat{j}+\widehat{k} \right)$
(B) $-\left( 3\widehat{i}+5\widehat{j}+3\widehat{k} \right)$
(C) $-\left( 6\widehat{i}+5\widehat{j}+2\widehat{k} \right)$
(D) $-\left( 2\widehat{i}+3\widehat{j}+\widehat{k} \right)$

Answer 1

Hint- Without some detailed knowledge of what created the field, knowledge of the importance of the electric field at a point is all that is required to decide what will happen to electric charges near to that specific point.

Complete step by step solution
Electric field, an electric property associated with each space point when some form of charge is present. The value of E, called electric field power or electric field force, or simply the electric field, reflects the magnitude and position of the electric field.
Instead of treating electrical force as a direct impact between two electrical charges at a distance from each other, the source of an electrical field expanding outward into the ambient space is assumed to be one charge, and the force imposed in that space on a second charge is thought to be a direct interaction between the electrical field and the second charge.
Electric field at any point can be written as
\[E=\dfrac{F}{q}\]
Now, Let us see what Cartesian Coordinate is-
A Cartesian Coordinate System is a system that uses a series of numerical coordinates to show a given point identified in a plane. The coordinates represent the distance of the point from two fixed perpendicular lines, referred to as axes (axis plural). In the same unit of length, all axes measure points and stretch over the actual n-space.
In Cartesian coordinate system any point in a spatial position is represented by x y and z that is we write it as \[P\left( x,y,z \right)\]
One more thing that we have is vectors in Cartesian coordinates represented manually as three perpendicular vectors that is \[\widehat{i},\widehat{j},\widehat{k}\]
Now, we are aware of the three points of the Cartesian system that are x, y and z and the formula for electric field is also known as discussed above.
For finding the electric field of each coordinate we need to differentiate the potential given with respect to that coordinate-
After looking at the above concepts we can collectively say that-
$\overrightarrow{E}=\left[ \dfrac{\partial V}{\partial x}\widehat{i}+\dfrac{\partial V}{\partial y}\widehat{j}+\dfrac{\partial V}{\partial z}\widehat{k} \right]$
Now in the question the values of V are given in terms of x, y, z simply substitute it in the above equation.
${{\overrightarrow{E}}_{{}}}=-\left[ \dfrac{\partial \left( 6xy-y+2yz \right)}{\partial x}\widehat{i}+\dfrac{\partial \left( 6xy-y+2yz \right)}{\partial y}\widehat{j}+\dfrac{\partial \left( 6xy-y+2yz \right)}{\partial z}\widehat{k} \right]$
$\Rightarrow \overrightarrow{E}=-\left[ (6y)\widehat{i}+(6x-1+2z)\widehat{j}+(2y)\widehat{k} \right]$
Now, in the question it is given that we have to find electric field at point \[\left( 1\text{ },\text{ }1\text{ },\text{ }0 \right)\],
So the next step is to put the values $x=1,y=1,z=0$
On pitting the values we –
$\Rightarrow \overrightarrow{E}=-\left[ (6\times 1)\widehat{i}+(6\times 1-1+2\times 0)\widehat{j}+(2\times 1)\widehat{k} \right]$
$\Rightarrow \overrightarrow{E}=-\left[ 6\widehat{i}+5\widehat{j}+2\widehat{k} \right]$

Thus, the correct choice for the given question is option C.

Note: The plane is divided into four infinite regions known as quadrants by a regular two dimensional Cartesian Coordinate System. Quadrants are defined in a counter-clockwise fashion, often denoted by Roman numerals from 1 to 4, where the first quadrant is defined as the upper right-hand space. Therefore, the plane is separated by a three-dimensional structure into not four, but eight zones, known as octants.