Question

The electric potential V at any point P(x,y,z) in space is given by V= 4$x^2$ V. The electric field at the point (1m, 2m) is:
A. -8i
B. 8i
C. -16i
D. 16i

Answer 1

Hint: We will use the formula of electric potential. As we need to find electric fields at a given point we will apply the vector form of electric potential. On differentiating the potential given in the question we can determine the electric field.

Formula used: Electric potential \[\vec{E}=-\dfrac{\partial V}{\partial x}i-\dfrac{\partial V}{\partial y}j-\dfrac{\partial V}{\partial z}k\]

Complete answer:
At point P the electric potential is given. We need to find the electric field at (1,2).
Electric field is an area that surrounds electric charge and has influence on all other charges present in the field. It is a vector quantity. We can also define electric potential as the rate of change of potential with respect to its direction.
The amount of work done to bring a charge from infinity to a point in the electric field is called electric potential. It is a scalar quantity.
The relation between electric potential and electric field is given by \[E=-\nabla V\]
Where E is the electric field and V is the electric potential.
In vector form it can be written as
\[\vec{E}=-\dfrac{\partial V}{\partial x}i-\dfrac{\partial V}{\partial y}j-\dfrac{\partial V}{\partial z}k\]
Given \[V=4{{x}^{2}}V\]
On differentiating equation (1) we will get the electric field
Here \[\dfrac{\partial V}{\partial x}=8x\]
\[\dfrac{\partial V}{\partial y}=\dfrac{\partial V}{\partial z}=0\]
Therefore \[\vec{E}=-8xi\]
At point (1,2) we need to find the electric field which means x=1 and y=2
Hence, \[\vec{E}=-8xi\]
Electric field at point (1,2) is found to be \[-8iN/C\] .

So option A is correct.

Note:
Electric fields can also be defined as the gradient of potential. Direction is taken positive when field is directed from lower potential to higher potential, whereas direction is taken negative when field is directed from higher potential to lower potential.