JEE MAIN - Physics (2009 - No. 12)
Two points $$P$$ and $$Q$$ are maintained at the potentials of $$10$$ $$V$$ and $$-4$$ $$V$$, respectively. The work done in moving $$100$$ electrons from $$P$$ to $$Q$$ is :
$$9.60 \times {10^{ - 17}}J$$
$$ - 2.24 \times {10^{ - 16}}J$$
$$ 2.24 \times {10^{ - 16}}J$$
$$- 9.60 \times {10^{ - 17}}J$$
Açıklama
$$ {{{W_{PQ}}} \over q} = \left( {{V_Q} - {V_P}} \right)$$
$$ \Rightarrow {W_{PQ}} = q\left( {{V_Q} - {V_P}} \right)$$
$$ = \left( { - 100 \times 1.6 \times {{10}^{ - 19}}} \right)\left( { - 4 - 10} \right)$$
$$ = + 2.24 \times {10^{ - 16}}J$$
$$ \Rightarrow {W_{PQ}} = q\left( {{V_Q} - {V_P}} \right)$$
$$ = \left( { - 100 \times 1.6 \times {{10}^{ - 19}}} \right)\left( { - 4 - 10} \right)$$
$$ = + 2.24 \times {10^{ - 16}}J$$