Question

An air-filled parallel-plate capacitor with plate separation \(d\) and plate area \(A\) is connected to a battery that applies a voltage \(V_{0}\) between plates. With the battery left connected, the plates are moved apart to a distance of \(10 d\). Determine by what factor each of the following quantities changes: \((a) V_{0} ;(b) C ;(c) E ;(d) D ;(e) Q ;(f) \rho_{S} ;(g) W_{E}\)

Short answer

When the plate separation of an air-filled parallel-plate capacitor connected to a constant-voltage battery increases from \(d\) to \(10d\), the following factors change: (a) Voltage (\(V_0\)): Remains unchanged (factor of 1). (b) Capacitance (\(C\)): Decreases by a factor of 10. (c) Electric field (\(E\)): Decreases by a factor of 10. (d) Electric displacement field (\(D\)): Decreases by a factor of 10. (e) Charge (\(Q\)): Decreases by a factor of 10. (f) Surface charge density (\(\rho_S\)): Decreases by a factor of 10. (g) Energy stored in the capacitor (\(W_E\)): Decreases by a factor of 10.

Step-by-step solution

Voltage (\(V_0\))

The battery maintains a constant voltage (\(V_0\)) between the plates. So, when the plates are moved apart, the voltage between the plates does not change because the battery is still connected. Therefore, the factor by which the voltage changes is 1.

Capacitance (\(C\))

The capacitance formula for a parallel-plate capacitor is \(C = \frac{\epsilon_0 A}{d}\). When the plate separation is increased to \(10d\), the new capacitance \(C'\) will be: $$C' = \frac{\epsilon_0 A}{10d}.$$ Now, we can find the factor by which capacitance changes by dividing \(C'\) by \(C\): $$\frac{C'}{C} = \frac{\frac{\epsilon_0 A}{10d}}{\frac{\epsilon_0 A}{d}} = \frac{1}{10}.$$ The capacitance decreases by a factor of 10.

Electric field (\(E\))

The electric field between the plates of a charged capacitor is given by \(E=\frac{V}{d}\), where \(V\) is the voltage across the plates. Since the voltage remains constant and the plate separation increases to \(10d\), the new electric field \(E'\) will be: $$E'=\frac{V_0}{10d}.$$ To find the factor by which the electric field changes, we can divide \(E'\) by \(E\): $$\frac{E'}{E}=\frac{\frac{V_0}{10d}}{\frac{V_0}{d}}=\frac{1}{10}.$$ The electric field decreases by a factor of 10.

Electric displacement field (\(D\))

The electric displacement field between the plates of a capacitor in a vacuum is given by \(D=\epsilon_0 E\), where \(\epsilon_0\) is the vacuum permittivity and \(E\) is the electric field. Since the electric field decreases by a factor of 10, the electric displacement field will also decrease by the same factor. Thus, the factor by which the electric displacement field changes is \(\frac{1}{10}\).

Charge (\(Q\))

The charge on the capacitor is given by \(Q=CV\) for a capacitor with a constant voltage. Since the capacitance decreases by a factor of 10, and the voltage remains constant, the charge will also decrease by a factor of 10. Thus, the factor by which the charge changes is \(\frac{1}{10}\).

Surface charge density (\(\rho_S\))

The surface charge density on the plates of a capacitor is given by \(\rho_S=\frac{Q}{A}\), where \(Q\) is the charge and \(A\) is the area of the plates. Since the charge decreases by a factor of 10 and the area remains constant, the surface charge density will also decrease by a factor of 10. Thus, the factor by which the surface charge density changes is \(\frac{1}{10}\).

Energy stored in the capacitor (\(W_E\))

The energy stored in the capacitor is given by \(W_E=\frac{1}{2}CV^2\). Since the capacitance decreases by a factor of 10 and the voltage remains constant, the energy stored will also decrease by a factor of 10. Thus, the factor by which the energy stored changes is \(\frac{1}{10}\). In summary: - The voltage \(V_0\) remains unchanged (factor of 1). - The capacitance \(C\) decreases by a factor of 10. - The electric field \(E\) decreases by a factor of 10. - The electric displacement field \(D\) decreases by a factor of 10. - The charge \(Q\) decreases by a factor of 10. - The surface charge density \(\rho_S\) decreases by a factor of 10. - The energy stored in the capacitor \(W_E\) decreases by a factor of 10.