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2. Electric Potential and Capacitance
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Two positive point charges of $12\,\mu C$ and $8\,\mu C$ are $10\,cm$ apart. The work done in bringing them $4\, cm$ closer is
A
$5.8\, J$
B
$5.8 \,eV$
C
$13 \,J$
D
$13 \,eV$
Solution
(c) $W = {U_f} – {U_i} = 9 \times {10^9} \times {Q_1}\,{Q_2}\left[ {\frac{1}{{{r_2}}} – \frac{1}{{{r_1}}}} \right]$
$==>$ $W = 9 \times {10^9} \times 12 \times {10^{ – 6}} \times 8 \times {10^{ – 6}}\left[ {\frac{1}{{4 \times 10{\,^{ – 2}}}} – \frac{1}{{10\, \times 10{\,^{ – 2}}}}} \right]$ $= 12.96 \,J ≈ 13 \,J$
Standard 12
Physics
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