Question

a certain capacitor has a capacitance of 2.00 μf. after it is charged to 10.0 μc and is isolated, the plates are brought closer together so its capacitance becomes 8.00 μf. what is the work done by the agent?

Explanation:

Step1: Recall energy formula for capacitor

The energy stored in a capacitor is $U = \frac{Q^{2}}{2C}$, where $Q$ is the charge and $C$ is the capacitance. The work - done by the agent is equal to the change in the energy of the capacitor.

Step2: Calculate initial energy $U_1$

Given $C_1 = 2.00\ \mu F=2\times 10^{- 6}\ F$ and $Q = 10.0\ \mu C = 10\times10^{-6}\ C$. Using $U_1=\frac{Q^{2}}{2C_1}$, we have $U_1=\frac{(10\times 10^{-6})^{2}}{2\times2\times 10^{-6}}=\frac{100\times10^{- 12}}{4\times10^{-6}} = 2.5\times10^{-5}\ J$.

Step3: Calculate final energy $U_2$

When $C_2 = 8.00\ \mu F=8\times 10^{-6}\ F$ and $Q$ remains the same ($Q = 10\times10^{-6}\ C$). Using $U_2=\frac{Q^{2}}{2C_2}$, we get $U_2=\frac{(10\times 10^{-6})^{2}}{2\times8\times 10^{-6}}=\frac{100\times10^{-12}}{16\times10^{-6}}=6.25\times 10^{-6}\ J$.

Step4: Calculate work - done $W$

The work - done $W=\Delta U=U_2 - U_1$. So $W=6.25\times 10^{-6}-2.5\times10^{-5}=- 1.875\times10^{-5}\ J$. The negative sign indicates that the agent does negative work (the electric field does positive work in bringing the plates closer). The magnitude of the work done by the agent is $|W| = 1.875\times10^{-5}\ J$.

Answer:

$1.875\times 10^{-5}$