An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d. When the capacitor plates carry charges + Q and - Q, the capacitor stores energy U 0. If the separation between the plates is doubled, how much electrical energy is stored in the capacitor?

2dQ²/2εA = 2U₀

Explanation:

The energy stored in the ideal parallel plate capacitor U = Q²/2C. For a parallel plate capacitor, C = εA/d where A is the area between the plates and D is the distance between them. So

U = Q²/2C = Q²/2εA/d = dQ²/2εA. At distance d₁ = d, U = U₀,

U₀ = dQ²/2εA,

When d₂ = 2d, U₁ = d₂Q²/2εA = 2dQ²/2εA = 2U₀

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg