Courses for Kids
Free study material
Offline Centres
Store Icon

The potential at a point x(measured in $\mu m$) due to some charges situated on the x-axis is given by V(x) $ = \dfrac{{20}}{{{x^2} - 4}}$ volt. The electric field E at $x = 4\mu m$ is given by:
A) $(10/9)volt/ \mu m$ and in the + ve x direction
B) $(5/3) volt/ \mu m$ and in the - ve x direction
C) $(5/3) volt/ \mu m$ and in the + ve x direction
D) $(10/9)volt/ \mu m$ and in the - ve x direction

Last updated date: 13th Jul 2024
Total views: 61.2k
Views today: 1.61k
61.2k+ views
Hint: We have the relationship between electric field and potential difference which is given by:
$E = \dfrac{V}{d}$. Where E is the electric field of the charge present, V is the potential difference of the cell, d is the length of balance for E.
Where electric field is given when change in potential takes place by:
$E = - \dfrac{{dV}}{{dx}}$(Minus sign shows the opposite direction from potential difference)
Using the above relations we will solve the problem.

Complete step by step solution:
Let’s define the electric field and potential difference first.
Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outwards from the positive charge and radially inwards from the negative charge. Electric field is generated around positive or negative charge, field may be attracting or repelling.
Potential difference: Potential difference: potential difference is the difference in electric potential between two points, which is defined as the work needed per unit of charge to move a test charge between the two points.
Now comes the calculation part:
$V(x) = \dfrac{{20}}{{({x^2} - 4)}}$ (Potential difference)
Where electric field is given as,
$E = - \dfrac{{dV}}{{dx}}$ (negative sign indicates direction of electric field is opposite to the direction of potential difference)
Let’s do the differentiation of potential differences.
$ \Rightarrow E = - \dfrac{{d\dfrac{{20}}{{({x^2} - 4)}}}}{{dx}}$
$ \Rightarrow E = - \dfrac{{ - 40x}}{{{{({x^2} - 4)}^2}}}$
$ \Rightarrow E = \dfrac{{40x}}{{{{\left( {{x^2} - 4} \right)}^2}}}$ ...............(1)
Value of x is given as $x = 4\mu m$
$ \Rightarrow E = \dfrac{{40 \times 4}}{{{{\left( {{4^2} - 4} \right)}^2}}} \\
   \Rightarrow E = \dfrac{{10}}{9}volt/\mu m \\$ (We have substituted the value of x in equation 1)
Electric field is positive which means the positive direction is in the positive direction of x.

Thus option (A) is correct.

Note: Electric field has many applications which we observe in our daily life such used in Van de graaff generators, Xerography is a dry copying process based on electrostatics, laser printers, inkjet printers and many other uses in medical fields such as MRI machines.