JEE Main & Advanced
JEE Main Paper (Held on 08-4-2019 Afternoon)
question_answer
A parallel plate capacitor has\[1\mu F\] capacitance. One of its two plates is given\[+2\mu C\]charge and the other plate, \[+4\mu C\] charge. The potential difference developed across the capacitor is:- [JEE Main 8-4-2019 Afternoon]
A)5V
B)2V
C)3V
D) 1V
Correct Answer:
D
Solution :
Charges at inner plates are \[1\mu C\]and \[1\mu C\] \[\therefore \]Potential difference across capacitor \[=\frac{q}{c}=\frac{1\mu C}{1\mu F}=\frac{1\times {{10}^{-6}}C}{1\times {{10}^{-6}}Farad}=1V\]