In moving from $A$ to $B$ along an electric field line, the electric field does $6.4 \times {10^{ - 19}}\,J$ of work on an electron. If ${\phi _1},\;{\phi _2}$ are equipotential surfaces, then the potential difference $({V_C} - {V_A})$ is.....$V$
Medium
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(b) Work done by the field $W = q( - dV) = - e({V_A} - {V_B})$
= $e({V_B} - {V_A})$ = $e({V_C} - {V_A})$ $(\because V_B = V_C)$
$==>$ $({V_C} - {V_A})$ $ = \frac{W}{e} = \frac{{6.4 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}} = 4\,V$
= $e({V_B} - {V_A})$ = $e({V_C} - {V_A})$ $(\because V_B = V_C)$
$==>$ $({V_C} - {V_A})$ $ = \frac{W}{e} = \frac{{6.4 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}} = 4\,V$
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