Calculating the Capacitance — Question 4

Question 4

A floating metal slab removes voltage drop from part of a plate gap

A neutral conducting slab of thickness t is placed between parallel plates of area A and separation d, with 0<t<d. It touches neither plate. The remaining vacuum gaps have widths g1 and g2. Neglect fringing.

1. Find the gap fields and capacitance.

2. Determine whether moving the slab sideways along the gap normal changes C, provided it remains fully overlapping and out of contact.

Question 4 - Solution

A floating metal slab removes voltage drop from part of a plate gap

(a) Method and calculation

$$g_1+g_2=d-t,\quad E_{\mathrm{metal}}=0,\quad E_1=E_2=Q/(\epsilon_0A).$$$$V=E_1g_1+E_2g_2=\frac{Q(d-t)}{\epsilon_0A},\quad C=\frac{\epsilon_0A}{d-t}.$$

(b) Interpretation and further calculation

$$g_1\to g_1+\delta,\quad g_2\to g_2-\delta\Rightarrow\Delta(g_1+g_2)=0.$$$$C\prime=C,\quad\frac{C}{C_{\mathrm{empty}}}=\frac d{d-t}>1.$$

Physical check

Neutrality forces equal induced charge magnitudes on the slab faces; its position does not change the total vacuum thickness.