A capacitor with capacitance 5μF is charged to 5μC. If the plates are pulled apart to reduce the capacitance to 2μF, how much work is done ?

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:-

Voltage rating of a parallel plate capacitor is 500V. Its dielectric can withstand a maximum electric field of 10⁶ V/m. The plate area is 10^–4 m². What is the dielectric constant is the capacitance is 15 pF? (given $$\varepsilon $$₀ = 8.86 × 10^–12 C²/Nm²)

A parallel plate capacitor with plates of area 1 m² each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is :

(Take $$\varepsilon $$₀ = 8.85 $$ \times $$ 10^$$-$$12 $${{{C^2}} \over {N - {m^2}}}$$)