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 :

Correct option is D. 1V∵Electric fd. due to cap plate is σ/2ε0Since both the plate has given positive charges. E=σ/2ε0Net Electric .field. σ/2/2ε0 - σ/1/2ε0 σ=Q/A Enet=Q/2/2AE0 - Q/1/2AE0 Enet=1/1AE0(Q2-Q1) here Q2=4μc Here Q2=2μc Enet=1/2AE0(4-2) =1/AE0 Also V=ED=Enet × d=1/AE0 × d=1/c 1∵C=E0A/d ∵C=1μF ∴V=1V V=Q/C=1μC/1μF=1Volt