Some equipotential surfaces are shown in the | vedclass

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

Question

(A) The electric field $E$ is related to the potential difference $\Delta V$ and the perpendicular distance $d$ between equipotential surfaces by the

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(A) The electric field $E$ is related to the potential difference $\Delta V$ and the perpendicular distance $d$ between equipotential surfaces by the formula $E = \frac{\Delta V}{d}$.From the figure,the potential difference between two adjacent equipotential surfaces is $\Delta V = 20 \ V - 10 \ V = 10 \ V$.The distance along the $x$-axis between these two surfaces is $\Delta x = 20 \ cm - 10 \ cm = 10 \ cm = 0.1 \ m$.The perpendicular distance $d$ between the surfaces is given by $d = \Delta x \sin 30^{\circ} = 0.1 	imes \sin 30^{\circ} = 0.1 	imes 0.5 = 0.05 \ m$.Therefore,the magnitude of the electric field is $E = \frac{10 \ V}{0.05 \ m} = 200 \ V/m$.

Some equipotential surfaces are shown in the figure below. The magnitude of the electric field will be $:-$ (in $V/m$)

Similar Questions

Variation in electric potential is maximum if one goes

In a space having electric field $\vec{E}=A(x \hat{i}+y \hat{j})$,the potential at a point $(10 \ m, 20 \ m)$ is zero. Then the potential at the origin is $\left[A=10 \ Vm^{-2}\right]$. (in $V$)

The electric potential at any point $(x, y, z)$ in metres is given by $V = 3x^2$. The electric field at a point $(2, 0, 1)$ is (in $Vm^{-1}$)

The variation of electric potential $V$ with distance $x$ from a fixed point is shown in the figure. What is the value of the electric field at $x = 2 \ m$?

The electric potential $V$ at any point $(x, y, z),$ all in metres in space is given by $V = 4x^2$ volt. The electric field at the point $(1, 0, 2)$ in volt/meter,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module