JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 14)
In the figure shown below, the charge on the left plate of the 10$$\mu $$F capacitor is –30$$\mu $$C. The charge on the right plate of the 6 $$\mu $$F capacitor is :
_11th_January_Morning_Slot_en_14_1.png)
+ 12 $$\mu $$C
+ 18 $$\mu $$C
$$-$$ 18 $$\mu $$C
$$-$$ 12 $$\mu $$C
Explanation
_11th_January_Morning_Slot_en_14_2.png)
6$$\mu $$F & 4$$\mu $$F are in parallel & total charge on this combination is 30 $$\mu $$C
$$ \therefore $$ Charge on 6$$\mu $$F capacitor = $${6 \over {6 + 4}} \times 30$$
= 18 $$\mu $$C
Since charge is asked on right plate therefore is +18$$\mu $$C
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