In moving from $\mathrm{A}$ to $\mathrm{B}$ along an electric field line, the work done by the electric…

In moving from $\mathrm{A}$ to $\mathrm{B}$ along an electric field line, the work done by the electric field on an electron is $6.4 \times 10^{-19} \mathrm{~J}$. If $\phi_{1}$ and $\phi_{2}$ are equipotential surfaces, then the potential difference $\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{A}}$
is
  1. $-4 \mathrm{~V}$
  2. $4 \mathrm{~V}$
  3. zero
  4. $6.4 \mathrm{~V}$

Solution

$\mathrm{W}_{\text {el. }}=\mathrm{q}\left(\mathrm{V}_{\mathrm{i}}-\mathrm{V}_{\mathrm{f}}\right)$
or $6.4 \times 10^{-19}=-1.6 \times 10^{-19}\left(\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}\right)$
or $\quad V_{A}-V_{B}=-4 V$
or $\quad \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}=-4 \mathrm{~V} \quad\left(\because \mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{C}}\right)$
or $\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{A}}=4 \mathrm{~V}$ *

Asked in: JEE Mains - Electrostatics - Test 3

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