University

Physics 2

Physics 2

Fall 2024

A parallel-plate capacitor with plate separation \(d\) is connected to a battery that has potential difference \(\Delta V_{\rm{bat}}\).

Without breaking any of the connections, insulating handles are used to increase the plate separation to \(2d\).

- Does the potential difference \(\Delta V_{C}\) change as the separation increases? If so, by what factor?
- Does the capacitance change? If so, by what factor?
- Does the capacitor charge \(Q\) change? If so, by what factor?

- No. The potential difference across the capacitor is determined by \(\Delta V_{\rm{bat}}\), which does not change.
- Yes. \(C = \epsilon_0 A / d\), so \(C\) decreases by a factor of 2.
- Yes. \(C = Q/\Delta V_{C}\) with no change in \(\Delta V_{C}\). So \(Q\) decreases by a factor of 2.

Each capacitor in the networks below has capacitance \(C\). What is the equivalent capacitance of the whole group?

- \(C/2\)
- \(2C\)
- \(C/3\)
- \(3C\)
- \(C\)
- \(2C/5\)

Rank in order, from largest to smallest, the equivalent capacitances \((C_{\rm{eq}})_1\) to \((C_{\rm{eq}})_4\) of these four groups of capacitors.

\((C_{\rm{eq}})_2 > (C_{\rm{eq}})_4 > (C_{\rm{eq}})_3 > (C_{\rm{eq}})_1\)

\((C_{\rm{eq}})_1 = \frac{1}{3} C\)

\((C_{\rm{eq}})_2 = 3 C\)

\((C_{\rm{eq}})_3 = \frac{2}{3} C\)

\((C_{\rm{eq}})_4 = \frac{3}{2} C\)

The gap between two capacitor plates is partially filled with a dielectric.

Rank the electric field strengths \(E_1\), \(E_2\), and \(E_3\) at points 1, 2, and 3.

The dielectric reduces the field in it.

\(E_1 = E_3 > E_2\)

When a \(360\) nF air capacitor is connected to a power supply, the energy stored in the capacitor is \(18.5\) μJ. While the capacitor is connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by \(23.2\) μJ.

- What is the potential difference between the capacitor plates?
- What is the dielectric constant of the slab?

- \(10.1\) V
- \(1.25\)

An air-filled (empty) parallel-plate capacitor is made from two square plates that are \(25\) cm on each side and \(1.0\) mm apart. The capacitor is connected to a \(50\) V battery and fully charged. It is then disconnected from the battery and its plates are pulled apart to a separation of \(2.00\) mm.

- What is the capacitance of this new capacitor?
- What is the charge on each plate?
- What is the potential difference between the plates after the adjustment?
- What is the electrical field between the plates before and after the adjustment?

- \(0.277\) nF
- \(27.7\) nC (before and after the adjustment)
- \(V = Q/F = 100\) V
- \(50\) kV/m (before and after the adjustment)

Before adjustment, \(C_0 = \epsilon_0 A/d = 5.53 \times 10^{-10}\) F.

For electric field, recall that \(E\) above an infinite plane charge (which is the approximation we’re making for the parallel plate capacitor) is independent of distance.

The network of capacitors shown below are all uncharged when a \(300\) V potential is applied between points \(A\) and \(B\) with the switch \(S\) open. (With \(B\) connected to ground, consider it to have potential \(V_B=0\) V.)

- What is the potential difference \(V_E − V_D\)?
- What is the potential at point \(E\) after the switch is closed?

- 100 V
- 150 V

For parts a and b, find the equivalent capacitance of the combined capacitors, then use \(C=Q/V\) (and its variations) to get the voltages.

*Last modified: Fri March 01 2024, 11:16 AM.*