Equipotential surfaces are given as shown. Th | vedclass

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(D) The relation between electric field $E$ and potential $V$ is given by $E = -\frac{dV}{dx}$.From the given figure,the potential increases in the po

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$- 1000 \,\, \hat{i} \,\, V/cm$

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(D) The relation between electric field $E$ and potential $V$ is given by $E = -\frac{dV}{dx}$.From the given figure,the potential increases in the positive $x$-direction.The change in potential $\Delta V = 10 \,\, V - 0 \,\, V = 10 \,\, V$.The change in distance $\Delta x = 1 \,\, cm - 0 \,\, cm = 1 \,\, cm = 10^{-2} \,\, m$.The magnitude of the electric field is $E = \frac{\Delta V}{\Delta x} = \frac{10 \,\, V}{1 \,\, cm} = 10 \,\, V/cm$.Since the electric field always points from higher potential to lower potential,and here the potential increases as $x$ increases,the electric field must point in the negative $x$-direction.Therefore,$\vec{E} = -10 \,\, \hat{i} \,\, V/cm = -1000 \,\, \hat{i} \,\, V/m$.

Equipotential surfaces are given as shown. The electric field intensity is ........

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