Question

A 20.0-\(\mu\)F capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to those of an uncharged 10.0-\(\mu\)F capacitor. Compute (a) the original charge of the system, (b) the final potential difference across each capacitor, (c) the final energy of the system, and (d) the decrease in energy when the capacitors are connected.

Short answer

Original charge is 0.016 C. Final voltage is 533.33 V. Final energy is 4.2667 J. Decrease in energy is 2.1333 J.

Step-by-step solution

Calculate the Original Charge

The original charge on the 20.0-\(\mu\)F capacitor is calculated using the formula \(Q = C \times V\). Here, \(C = 20.0\,\mu\text{F}\) and \(V = 800\,\text{V}\). Therefore, \(Q = 20.0 \times 10^{-6}\,\text{F} \times 800\,\text{V} = 0.016\,\text{C}\).

Calculate the Final Potential Difference

The total charge is conserved, so the charge \(Q = 0.016\,\text{C}\) is shared between the two capacitors connected in parallel. The equivalent capacitance \(C_{eq}\) is the sum of the two capacitances: \(C_{eq} = 20.0\,\mu\text{F} + 10.0\,\mu\text{F} = 30.0\,\mu\text{F}\). The final voltage \(V_f\) across both capacitors is given by \(V_f = \frac{Q}{C_{eq}} = \frac{0.016\,\text{C}}{30.0 \times 10^{-6}\,\text{F}} = 533.33\,\text{V}\).

Calculate the Final Energy of the System

The final energy stored in the system is given by \(E_f = \frac{1}{2} C_{eq} V_f^2\). Substitute \(C_{eq} = 30.0 \times 10^{-6}\,\text{F}\) and \(V_f = 533.33\,\text{V}\) to get \(E_f = \frac{1}{2} \times 30.0 \times 10^{-6}\,\text{F} \times (533.33)^2\,\text{V}^2 = 4.2667\,\text{J}\).

Calculate the Original Energy

The original energy stored in the 20.0-\(\mu\)F capacitor is \(E_i = \frac{1}{2} C V^2 = \frac{1}{2} \times 20.0 \times 10^{-6}\,\text{F} \times (800)^2\,\text{V}^2 = 6.4\,\text{J}\).

Calculate the Decrease in Energy

The decrease in energy when the capacitors are connected is the difference between the original energy and the final energy: \(E_{ ext{decrease}} = E_i - E_f = 6.4\,\text{J} - 4.2667\,\text{J} = 2.1333\,\text{J}\).

Key concepts

Electrostatics

Electrostatics is the branch of physics that deals with the study of forces, fields, and potentials arising from static electric charges. In this context, we talk about capacitors, devices used to store electrical energy in an electric field. Capacitors have two conductors separated by an insulator, and when connected to a voltage source, such as a battery, they become charged.

When you consider a charged capacitor, like the initial 20.0-\(\mu\text{F}\) capacitor charged to 800 V, it holds a charge calculated by the formula \(Q = C \times V\), where \(C\) is the capacitance and \(V\) the voltage. A charged capacitor has a potential difference across its plates and an electric field inside that exerts forces on any charge placed within the field.

In our problem, this charged capacitor is connected to an uncharged 10.0-\(\mu\text{F}\) capacitor. When the connection is made, charge redistribution occurs due to electrostatic forces, causing a new uniform potential difference across both capacitors.

Energy Conservation

Energy conservation is a fundamental principle in physics that states energy cannot be created or destroyed but can only change forms. In electrostatic systems, energy is conserved during charge redistribution. However, you might notice a drop in the system's potential energy when redistribution occurs.

Initially, the 20.0-\(\mu\text{F}\) capacitor had energy \(E_i = \frac{1}{2} C V^2\). When connected to another capacitor, the total stored energy appears to decrease. This happens because some energy is transferred to heat and electromagnetic radiation during the rapid redistribution of charge.
The calculated initial energy was 6.4 J. After redistribution, the final energy becomes 4.2667 J. The apparent loss of 2.1333 J exemplifies how some energy escapes as heat due to non-ideal conditions in real capacitors, demonstrating energy conversion rather than loss.

Capacitance Calculation

Understanding capacitance and its calculation is essential when dealing with capacitors. Capacitance \(C\) is a measure of a capacitor's ability to store charge per unit voltage, given by \(C = \frac{Q}{V}\).

In our scenario, two capacitors are connected in parallel after the initial capacitor is charged. The total capacitance \(C_{eq}\) is the sum of the capacitances of both capacitors. So, \(C_{eq} = 20.0\, \mu\text{F} + 10.0\, \mu\text{F} = 30.0\, \mu\text{F}\).

• This illustrates how total capacitance in parallel combinations can be found by simple addition.
• It shows that parallel capacitors share the total charge across them, leading to the same potential difference \(V_f\) across each.

The resultant final voltage \(V_f\) is calculated using conservation of charge and the formula \(V_f = \frac{Q}{C_{eq}}\). Thus, the understanding of parallel capacitance is vital for solving such problems accurately.