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Standard XII

Physics

Potential Gradient and Relation between Electric Field and Potential

A uniform ele...

Question

A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be a point on the x-axis at x $= + 1 c m$ and C be the point on the y-axis at y $= + 1 c m$ . Then, the potential at points A, B and C satisfy :

V_{A} < V_{B}

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V_{A} > V_{B}

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V_{A} < V_{C}

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V_{A} > V_{C}

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Solution

The correct option is B $V_{A} > V_{B}$
As $E = - \frac{d V}{d x}$ so potential will less near the electric field direction.
Thus, $V_{A} = V_{C} > V_{B} \Rightarrow V_{A} > V_{B}, V_{C} > V_{B}$

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A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be a point on the x-axis at x =+1cm and C be the point on the y-axis at y =+1cm. Then, the potential at points A, B and C satisfy :

The correct option is B VA>VBAs E=−dVdx so potential will less near the electric field direction.Thus, VA=VC>VB⇒VA>VB,VC>VB

A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be a point on the x-axis at x $= + 1 c m$ and C be the point on the y-axis at y $= + 1 c m$ . Then, the potential at points A, B and C satisfy :

The correct option is B $V_{A} > V_{B}$
As $E = - \frac{d V}{d x}$ so potential will less near the electric field direction.
Thus, $V_{A} = V_{C} > V_{B} \Rightarrow V_{A} > V_{B}, V_{C} > V_{B}$