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Potential Gradient and Relation between Electric Field and Potential

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?

A

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

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B

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

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C

E_{2} = E_{4} = E_{5} a n d E_{1} < E_{3}

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D

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

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Solution

The correct option is B $E_{1} = E_{3} = E_{5}$
As potential gradient is given by,
$E = \frac{- d V}{d x}$

Where, in region $1, 3$ and $5$ gradient are constant i.e. zero so, $E_{1} = E_{3} = E_{5}$

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Similar questions

E = E_{1} \cap E_{2} \cap E_{3} \cap E_{4} \cap E_{5}

P (E_{1}) = \frac{95}{100}, P (E_{2} / E_{1}) = \frac{94}{99}, P (E_{3} / (E_{1} \cap E_{2})) = \frac{93}{98}, P (E_{4} / E_{1} \cap E_{2} \cap E_{3}) = \frac{92}{97}

P (E_{5} / E_{1} \cap E_{2} \cap E_{3} \cap E_{4}) = \frac{91}{96}

P (E)

E = E_{1} E_{2} E_{3} E_{4} E_{5}

P (E_{1}) = \frac{95}{100}

P (E_{2} / E_{1}) = \frac{94}{99}

P (E_{1} / E_{1} E_{2}) = \frac{93}{98}

P (E_{4} / E_{1} E_{2} E_{3}) = \frac{92}{97}

P (E_{5} / E_{1} E_{2} E_{3} E_{4}) = \frac{91}{96}

P (E)

.

E_{1}, E_{2}, E_{3} a n d E_{4}

are the energies required to knock off the

1^{s t}, 2^{n d}, 3^{r d}

4^{t h}

electrons respectively from outermost orbit of the Carbon atom, choose the correct option:

E_{1}, E_{2}, E_{3} a n d E_{4}

are the energies required to knock off the 1st, 2nd, 3rd and 4th electron respectively from the outermost orbit of the Carbon atom, choose the correct option:

E_{1}, E_{2}, E_{3} a n d E_{4}

are the energies required to knock off the

1^{s t}, 2^{n d}, 3^{r d} a n d 4^{t h}

electrons (from the last shell) respectively from outermost orbit of the Carbon atom, choose the correct option:

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