The energy stored in a capacitor is given by:
\( U = \frac{1}{2} C V^2 \)
Differentiating with respect to time:
\( \frac{dU}{dt} = C V \frac{dV}{dt} \)
Substituting the given values:
\( C = 5 \times 10^{-6} \, \text{F}, \quad V = 4 \, \text{V}, \quad \frac{dV}{dt} = 0.6 \, \text{V/s} \)
\( \frac{dU}{dt} = (5 \times 10^{-6}) \cdot 4 \cdot 0.6 = 12 \times 10^{-6} \, \text{W} = 12\,\mu\text{W} \)
Thus, the solution is \( {12\,\mu\text{W}} \).
The energy stored in a capacitor is given by: \[ E = \frac{1}{2} C V^2 \] The rate at which energy is stored is: \[ \frac{dE}{dt} = C \times V \times \frac{dV}{dt} \] Given: - \( C = 5 \times 10^{-6} \, \text{F} \) - \( V = 4 \, \text{V} \) - \( \frac{dV}{dt} = 0.6 \, \text{V/s} \) Thus: \[ \frac{dE}{dt} = (5 \times 10^{-6}) \times 4 \times 0.6 = 12 \times 10^{-6} \, \text{W} = 12 \, \mu\text{W} \] Thus, the rate at which energy is stored is \(12\ µW\).
A parallel plate capacitor has two parallel plates which are separated by an insulating medium like air, mica, etc. When the plates are connected to the terminals of a battery, they get equal and opposite charges, and an electric field is set up in between them. This electric field between the two plates depends upon the potential difference applied, the separation of the plates and nature of the medium between the plates.