The work done in placing a charge of 8times 10-18- coulomb on a condenser of capacity 100 micro-farad is-3-2times 10-26- joule16times 10-32- joule4times 10-10- joule32times 10-32- joule

Question

Toppr Admin · Accepted Answer

Here the work done - energy stored in capacitor -Q22Cor- W-8-xD7-10-x2212-18-22-xD7-100-xD7-10-x2212-6-32-xD7-10-x2212-32-xA0-J-xA0-

The work done in placing a charge of $8 \times 10^{- 18}$ coulomb on a condenser of capacity $100$ micro-farad is:

$3.2 \times 10^{- 26}$ joule
$16 \times 10^{- 32}$ joule
$4 \times 10^{- 10}$ joule
$32 \times 10^{- 32}$ joule

Here the work done $=$ energy stored in capacitor $= \frac{Q^{2}}{2 C}$
or, $W = \frac{(8 \times 10^{- 18})^{2}}{2 \times 100 \times 10^{- 6}} = 32 \times 10^{- 32} J$

The work done in placing a charge of 8×10−18 coulomb on a condenser of capacity 100 micro-farad is:3.2×10−26 joule16×10−32 joule4×10−10 joule32×10−32 joule

Here the work done = energy stored in capacitor =Q22Cor, W=(8×10−18)22×100×10−6=32×10−32 J

The work done in placing a charge of $8 \times 10^{- 18}$ coulomb on a condenser of capacity $100$ micro-farad is:

$3.2 \times 10^{- 26}$ joule
$16 \times 10^{- 32}$ joule
$4 \times 10^{- 10}$ joule
$32 \times 10^{- 32}$ joule

Here the work done $=$ energy stored in capacitor $= \frac{Q^{2}}{2 C}$
or, $W = \frac{(8 \times 10^{- 18})^{2}}{2 \times 100 \times 10^{- 6}} = 32 \times 10^{- 32} J$