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Byju's Answer

Standard XII

Physics

Motional EMF in an Open Conductor

The figure gi...

Question

The figure gives the electric potential $V$ as a function of distance through five regions on x-axis. Which of the following is true for the electric field $E$ in these regions:

E_{1} > E_{2} > E_{3} > E_{4} > E_{5}

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E_{1} = E_{3} = E_{5}

and

E_{2} < E_{4}

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E_{2} = E_{4} = E_{5}

and

E_{1} < E_{3}

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E_{1} < E_{2} < E_{3} < E_{4} < E_{5}

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Solution

The correct option is B $E_{1} = E_{3} = E_{5}$ and $E_{2} < E_{4}$
As we know, $E = - \frac{d V}{d x} . . . . . . (1)$
So, from the figure, $E_{1} = E_{3} = E_{5} = 0$
As $E_{2}$ has positive slope and $E_{4}$ has negative slope,
According to $(1), E_{2} < E_{4}$

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Similar questions

The figure gives the electric potential $V$ as a function of distance through five regions on $x -$ axis. Which of the following is true for the electric field $E$ in these regions?

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The figure gives the electric potential V as a function of distance through five regions on x-axis. Which of the following is true for the electric field E in these regions:

The correct option is B E1=E3=E5 and E2<E4As we know, E=−dVdx......(1)So, from the figure,E1=E3=E5=0As E2 has positive slope and E4 has negative slope,According to (1),E2<E4

The figure gives the electric potential $V$ as a function of distance through five regions on x-axis. Which of the following is true for the electric field $E$ in these regions:

The correct option is B $E_{1} = E_{3} = E_{5}$ and $E_{2} < E_{4}$
As we know, $E = - \frac{d V}{d x} . . . . . . (1)$
So, from the figure, $E_{1} = E_{3} = E_{5} = 0$
As $E_{2}$ has positive slope and $E_{4}$ has negative slope,
According to $(1), E_{2} < E_{4}$