At which point is the electric field maximum?

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The electric field $E$ is related to the potential gradient by the relation $E = -\frac{dV}{dr}$.This implies that the electric field is inversely

Vedclass · Accepted Answer

$A$

Vedclass · Answer

(B) The electric field $E$ is related to the potential gradient by the relation $E = -\frac{dV}{dr}$.This implies that the electric field is inversely proportional to the distance between the equipotential surfaces,i.e.,$E \propto \frac{1}{dr}$.Looking at the provided equipotential lines,the distance between the equipotential surfaces is smallest near point $A$ (where the lines are most crowded).Since the lines are closest together at point $A$,the potential gradient is steepest,and therefore the electric field is maximum at point $A$.

Similar Questions

$A$ uniform electric field of magnitude $100 \ V/m$ in space is directed along the line $y = 3 + x$. Find the potential difference between point $A(3, 1)$ and $B(1, 3)$ in Volts.

The figure shows two equipotential lines in the $XY$ plane for an electric field. The scale is shown. The $X$-component $E_x$ and $Y$-component $E_y$ of the electric field in the space between the equipotential lines are respectively:

In the electric field of a point charge $+q$, a certain charge is carried from point $A$ to points $B, C, D$ and $E$ along the paths shown in the figure. Then the work done:

What is the direction of the lines of force at any point on the equipotential surface?

$A$ charge $+Q$ is placed at the centre of a circular path of radius $r$. The work done to bring a charge $+q$ from one end of the diameter to the other end of the circular path in the electric field produced by charge $+Q$ is $.......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module