Question

A 10 microfarad capacitor is connected to a 100V battery what is the electrostation energy stored

Answer 1

$Energy \: stored = \frac{1}{2} c \: {v}^{2} \\ \\ = \frac{1}{2} (10 \times {10}^{ - 6} ) {(100)}^{2} \\ = \frac{1}{2} \times {10}^{ - 5} \times {10}^{4} \\ = \frac{1}{20} joules$

$\huge{\red{\mathfrak{Solution}}}$⤴️⤴️

Answer 2

Answer:

E = 0.05 Joules

Explanation:

It is given that,

Capacitance of a capacitor, $C=10\ \mu C$

Voltage, V = 100 V

The electrostatic energy stored in the capacitor is given by :

$E=\dfrac{1}{2}CV^2$

$E=\dfrac{1}{2}\times 10\times 10^{-6}\times (100)^2$

E = 0.05 Joules

So, the electrostatic energy stored in the capacitor is 0.05 joules. Hence, this is the required solution.