JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 18)
A parallel plate capacitor has 1μF capacitance.
One of its two plates is given +2μC charge and
the other plate, +4μC charge. The potential
difference developed across the capacitor is:-
1V
5V
2V
3V
Explanation
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Charges at inner plates are 1 $$\mu $$C and –1 $$\mu $$C.$$ \therefore $$ Potential difference across capacitor
= $${q \over c} = {{1\mu C} \over {1\mu F}} = {{1 \times {{10}^{ - 6}}C} \over {1 \times {{10}^{ - 6}}Farad}} = 1V$$
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