
The electric potential decreases uniformly from 180V to 20V as one moves on the –axis from $x=-2cm$ to $x=+2cm$. the electric field at the origin:
A) Must be equal to $-40V/cm.$
B) Maybe equal to $40V/cm.$
C) Maybe less than $40V/cm.$
D) Maybe less than $-40V/cm.$
Answer
512.1k+ views
Hint: Here we will be using the relation between electric potential and electric field. Along the direction of the electric field, the potential is decreasing.
Formula used:
Electric field $E = - \dfrac{{dV}}{{dx}}$
Where $dV$ is the change in potential in the direction of the electric field $E$.
$dx$ is the perpendicular distance of the surface.
Complete step by step solution:
Given, the electric potential decreases uniformly, from $180V$ to $20V$.
That is change in potential is given by,
\[dV = {\text{ }}180 - 20 = 160V.\]
then, the perpendicular distance of the surface is from $-2cm$ to $+2cm$.
therefore, $dx = -2-(+2)=-4cm$.
now we have a equation,
Electric field $E = - \dfrac{{dV}}{{dx}}$
The negative sign indicates that the direction of $E$ is always in the direction of decrease of electric potential.
The electric field at the origin is given by,
$E = \dfrac{{ - \left( {180 - 20} \right)}}{{ - 2 - 2}}$
$E = \dfrac{{160}}{4} = 40V/cm.$ in the positive direction of X-axis.
Hence, 40V/cm is the electric field at the origin.
Additional information:
The electric potential at a point in an electric field is defined as the work done per unit positive test charge in bringing it from infinity to that point against the electric field. The field produced by the positive charge is directed away from the charge. So, in bringing a positive charge from infinity to any point in this field, we have to do work, and hence the potential at any point around a positive charge is positive. The potential difference between two points does not depend on the path followed. This is because the electrostatic field is a conservative field.
Note:
A positive charge gives rise to a positive potential. A negative charge produces negative potential.
An equipotential surface in an electric field is a surface on which the potential is the same at every point.
Formula used:
Electric field $E = - \dfrac{{dV}}{{dx}}$
Where $dV$ is the change in potential in the direction of the electric field $E$.
$dx$ is the perpendicular distance of the surface.
Complete step by step solution:
Given, the electric potential decreases uniformly, from $180V$ to $20V$.
That is change in potential is given by,
\[dV = {\text{ }}180 - 20 = 160V.\]
then, the perpendicular distance of the surface is from $-2cm$ to $+2cm$.
therefore, $dx = -2-(+2)=-4cm$.
now we have a equation,
Electric field $E = - \dfrac{{dV}}{{dx}}$
The negative sign indicates that the direction of $E$ is always in the direction of decrease of electric potential.
The electric field at the origin is given by,
$E = \dfrac{{ - \left( {180 - 20} \right)}}{{ - 2 - 2}}$
$E = \dfrac{{160}}{4} = 40V/cm.$ in the positive direction of X-axis.
Hence, 40V/cm is the electric field at the origin.
Additional information:
The electric potential at a point in an electric field is defined as the work done per unit positive test charge in bringing it from infinity to that point against the electric field. The field produced by the positive charge is directed away from the charge. So, in bringing a positive charge from infinity to any point in this field, we have to do work, and hence the potential at any point around a positive charge is positive. The potential difference between two points does not depend on the path followed. This is because the electrostatic field is a conservative field.
Note:
A positive charge gives rise to a positive potential. A negative charge produces negative potential.
An equipotential surface in an electric field is a surface on which the potential is the same at every point.
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