Correct Answer: Option D

Explanation

The energy stored in a given capacitor is given by
E = \(\frac{1}{2}\) CV\(^2\) = \(\frac{1}{2}\) QV = \(\frac{1}{2}\)\(\frac{Q^2}{\text{C}}\)
Where C = the capacitance of the capacitor
Q = charge carried by the capacitor
V = p.d across the plates of the capacitor, thus from the data given, C = 10μF
= 10 x 10\(^{-6}\)F
Q = 100μc = 100 x 10\(^{-6}\)C
∴ Energy stored = \(\frac{1}{2}\)\(\frac{Q^2}{\text{C}}\)

= \(\frac{1}{2}\)\(\frac{(100 \times 10^{-6})^2}{10 \times 10^{-6}}\)

= 0.0005 = 5 x 10\(^{-4}\)J

• What can i study with 155 score in science course? Answer this

  2 Answers   ·   Biology   ·   3 days ago
• Does WAEC give Calculator to student for there exam? Answer this

  3 Answers   ·   Mathematics   ·   4 days ago
• find the value of (16)^3/2 + log0.0001 + log325? Answer this

  13 Answers   ·   Mathematics   ·   10 years ago