JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 24)
A 2$$\mu$$F capacitor C1 is first charged to a potential difference of 10V using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor C2 of 8 $$\mu$$F. The charge in C2 on equilibrium condition is ____________ $$\mu$$C. (Round off to the Nearest Integer)
_17th_March_Evening_Shift_en_24_1.png)
উত্তর
16
ব্যাখ্যা
_17th_March_Evening_Shift_en_24_2.png)
When battery is removed & the capacitor is connected
2V + 8v = 20
10V = 20
V = 2 volt
$$ \because $$ Q = CV
Q = 8 $$\times$$ 2 = 16$$\mu$$c