(A) The relationship between electric field $E$ and electric potential $V$ is given by $E = -\frac{dV}{dx}$.From the given graph,it is observed that i

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(A) The relationship between electric field $E$ and electric potential $V$ is given by $E = -\frac{dV}{dx}$.From the given graph,it is observed that in the region between $x = 1 \ m$ and $x = 3 \ m$,the electric potential $V$ is constant $(V = 2 \ V)$.Since the potential is constant in this region,the rate of change of potential with respect to distance is zero,i.e.,$\frac{dV}{dx} = 0$.Therefore,the electric field at $x = 2 \ m$ is $E = -\frac{dV}{dx} = 0 \ V/m$.

Class 12 Physics Electrostatic Potential and Capacitance Relation between Electric Field and Potential and Potential Gradient

Medium

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 $x = 2 \ m$?

AP EAMCET 2020 All AP EAMCET

A
$0$
B
$3$
C
$2$
D
$6$

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Electrostatic Potential and Capacitance

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Relation between Electric Field and Potential and Potential Gradient

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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 $x = 2 \ m$?

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