A capacitor with capacitance 5 uF is charged to 5 uC. If the plates are pulled apart to reduce the c

Q: Why can't we use the formula $U = \frac{1}{2}CV^2$ to find the energy?

You can, but it's much harder! Since the capacitance $C$ changes and the charge $Q$ is constant, the voltage $V$ must also change ($V = Q/C$). Using $U = \frac{Q^2}{2C}$ avoids having to calculate the new voltage.

Q: Why does the potential energy increase when we pull the plates apart?

The plates have opposite charges, meaning they attract each other. To pull them apart, you must apply an external force and do positive work against this electrostatic attraction. This work is stored as additional potential energy.

Q: What happens to the voltage across the capacitor during this process?

Since $V = Q/C$ and the charge $Q$ is constant, decreasing the capacitance $C$ (by pulling the plates apart) actually causes the voltage $V$ to increase.

Q: Is the work done by the electric field positive or negative?

The work done by the electric field is negative. The electric force is attractive (pulling the plates together), but the displacement is outwards (pulling them apart). Since force and displacement are in opposite directions, the work is negative.

Q: What if the battery remained connected while pulling the plates apart?

If the battery was still connected, the voltage $V$ would remain constant. The charge $Q$ would decrease as capacitance decreases, and the battery would do work. The energy change would be calculated differently, using $U = \frac{1}{2}CV^2$.

Question

A capacitor with capacitance $5\mu	ext{F}$ is charged to $5\mu	ext{C}$. If the plates are pulled apart to reduce the capacitance to $2\mu	ext{F}$, how much work is done?

Options:

(A) $6.25 	imes 10^{-6}	ext{ J}$

(B) $2.16 	imes 10^{-6}	ext{ J}$

(C) $2.55 	imes 10^{-6}	ext{ J}$

(D) $3.75 	imes 10^{-6}	ext{ J}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a parallel plate capacitor that has been fully charged and then disconnected from its power source. We are given its initial capacitance as $C_1 = 5\mu	ext{F}$, and it holds a charge of $Q = 5\mu	ext{C}$.

Because the capacitor is no longer connected to a battery, the charge $Q$ is completely trapped on the plates. It has nowhere to go! This is a critical realization: throughout any physical changes we make to the capacitor, **the charge $Q$ remains strictly constant**.

### The Master Equation

We need to find the work done when the plates are pulled apart, reducing the capacitance to $C_2 = 2\mu	ext{F}$. According to the work-energy theorem, the work done by an external agent on the system is equal to the change in its potential energy, provided there are no other energy losses.

$$W = \Delta U = U_f - U_i$$

To calculate the potential energy, we have three equivalent formulas: $U = \frac{1}{2}CV^2$, $U = \frac{1}{2}QV$, and $U = \frac{Q^2}{2C}$. Since we established that the charge $Q$ is our constant anchor, the most elegant and straightforward formula to use is $U = \frac{Q^2}{2C}$. Using any other formula would force us to calculate the changing voltage, adding unnecessary complexity.

### Calculating the Work Done

Let's set up our master equation by substituting the energy formula for both the final and initial states:

$$W = \frac{Q^2}{2C_2} - \frac{Q^2}{2C_1}$$

We can factor out the common term $\frac{Q^2}{2}$ to make the algebra cleaner:

$$W = \frac{Q^2}{2} \left( \frac{1}{C_2} - \frac{1}{C_1} ight)$$

Now, we carefully substitute our given values. Remember to account for the 'micro' prefix, which stands for $10^{-6}$:

$$W = \frac{(5 	imes 10^{-6})^2}{2} \left( \frac{1}{2 	imes 10^{-6}} - \frac{1}{5 	imes 10^{-6}} ight)$$

### The Final Result

Let's simplify the expression step-by-step. Squaring the charge gives us $25 	imes 10^{-12}$. Inside the bracket, we can factor out $\frac{1}{10^{-6}}$:

$$W = \frac{25 	imes 10^{-12}}{2} 	imes \frac{1}{10^{-6}} \left( \frac{1}{2} - \frac{1}{5} ight)$$

This simplifies beautifully. The powers of ten combine to leave $10^{-6}$, and the fractions inside the bracket evaluate to $\frac{3}{10}$:

$$W = \frac{25 	imes 10^{-6}}{2} 	imes \left( \frac{3}{10} ight)$$

Multiplying the numerators and denominators gives $\frac{75}{20} 	imes 10^{-6}$, which perfectly reduces to our final answer:

$$W = 3.75 	imes 10^{-6}	ext{ J}$$

This positive work makes physical sense! The oppositely charged plates attract each other. To pull them apart, you must actively exert a force and do positive work against that electrostatic attraction, thereby increasing the stored potential energy of the system.

Animated Solution for Physics - Electrostatics: A capacitor with capacitance $5 μ F$ is charged to $5 μ C$ . If the plates are pulled apart to reduce the capacitance to $2 μ F$ , how much work is done?

Select Answer:

Visualized Solution

$C_{1}$ and $Q$

$U_{i} = \frac{Q ^{2}}{2 C _{1}}$

$C_{2} = 2 μ F$

$W = Δ U$

$W = \frac{Q ^{2}}{2} (\frac{1}{C _{2}} - \frac{1}{C _{1}})$

Substitution

Simplification

Final Answer

The Way Forward

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup

The Master Equation

Calculating the Work Done

The Final Result

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