Question bank/JEE Main/Physics

A 60 pF capacitor is fully charged by a 20V supply. It is then
disconnected from the supply and isconnected to another uncharged 60 pF
capacitor in parallel. The electrostatic energy that is lost in
thisprocess by the time the charge is redistributed between them is (in
nJ)________.
A rRi'pkr~ bls lzksr ls gVkdj 60 pF ds,d nwljs =k (parallel connection) esa tksM+k tc
esa forfjr gks tk;s rks bl
izfØ fLFkj oS|qr

A3B4C5D6

Show answer and solution▾

[IMAGE] V_0 = 20 V
Heat loss gkfu = U_i - U_f
$= \frac{1}{2}CV_{0}^{2} - 2\left\lbrack \frac{1}{2}C\left( \frac{V_{0}}{2} \right)^{2} \right\rbrack$
$= \frac{CV_{0}^{2}}{4}$
$= \frac{\left( 60 \times 10^{- 12} \right)(20)^{2}}{4}J$
= 6 × 10^-9 J = 6 nJ

Source: Grass-G / P-21 Physcis Paper 19 Dec Hindi Eng..docx