3) The amount of work done in increasing the voltage across the plates of the capacitor from \(5\text{ V}\) to \(10\text{ V}\) is \(W\). The work done in increasing it from \(10\text{ V}\) to \(15\text{ V}\) will be:
Solution:
The energy stored in a capacitor is given by: \[ U = \frac{1}{2}CV^2. \] The work done in increasing the voltage from \(V_1\) to \(V_2\) is the change in energy: \[ \Delta U = \frac{1}{2}C\left(V_2^2 - V_1^2\right). \] For the change from \(5\text{ V}\) to \(10\text{ V}\), we have: \[ W = \frac{1}{2}C\left(10^2 - 5^2\right) = \frac{1}{2}C(100 - 25) = \frac{1}{2}C \times 75. \] For the change from \(10\text{ V}\) to \(15\text{ V}\), the work done is: \[ W' = \frac{1}{2}C\left(15^2 - 10^2\right) = \frac{1}{2}C(225 - 100) = \frac{1}{2}C \times 125. \] Taking the ratio: \[ \frac{W'}{W} = \frac{\frac{1}{2}C \times 125}{\frac{1}{2}C \times 75} = \frac{125}{75} = \frac{5}{3} \approx 1.67. \] Thus, \(W' \approx 1.67W\), so option D is the correct answer.
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