Correct option is DGiven Data: C=5.0 μF=5×10−6 FC = 5.0 \,\mu F = 5 \times 10^{-6} \,FC=5.0μF=5×10−6F , V=100 VV = 100 \,VV=100VFormula for Energy Stored in a Capacitor:U=12CV2U = \frac{1}{2} C V^2U=21CV2U=12×5×10−6×(100)2U = \frac{1}{2} \times 5 \times 10^{-6} \times (100)^2U=21×5×10−6×(100)2U=12×5×10−6×104U = \frac{1}{2} \times 5 \times 10^{-6} \times 10^4U=21×5×10−6×104U=2.5×10−2 JU = 2.5 \times 10^{-2} \,JU=2.5×10−2J