Assertion (A): In a region of constant potent | vedclass

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(B) For Assertion $(A)$:The potential difference is given by $dV = -\vec{E} \cdot d\vec{r}$. If the electric field $\vec{E} = 0$,then $dV = 0$,which i

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Both $(A)$ and $(R)$ are true; $(R)$ is not the correct explanation of $(A)$

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(B) For Assertion $(A)$:The potential difference is given by $dV = -\vec{E} \cdot d\vec{r}$. If the electric field $\vec{E} = 0$,then $dV = 0$,which implies that the potential $V$ is constant throughout the region.For Reason $(R)$:According to Gauss's law,the net electric flux $\phi_E$ through a closed surface is given by $\phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}$.If $\vec{E} = 0$ everywhere in the region,then the flux $\phi_E = 0$,which implies $q_{enclosed} = 0$. Thus,the reason is correct.However,the reason explains why the charge must be zero if the field is zero,but it does not explain why the potential is constant in that region (which is derived from the relation $E = -
abla V$). Therefore,$(R)$ is not the correct explanation of $(A)$.

Assertion $(A)$: In a region of constant potential,the electric field is zero and there can be no charge inside the region.
Reason $(R)$: According to Gauss's law,the charge inside the region should be zero if the electric field is zero.

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Assertion $(A)$: In a region of constant potential,the electric field is zero and there can be no charge inside the region.Reason $(R)$: According to Gauss's law,the charge inside the region should be zero if the electric field is zero.