Four large parallel metal plates 1, 2, 3, and 4 are located at a small distance $d$ from one another

Question

Four large parallel metal plates 1, 2, 3, and 4 are located at a small distance $d$ from one another in that order. The extreme plates (1 and 4) are interconnected by means of a conducting wire. A potential difference $\Delta\phi$ is applied between the internal plates 2 and 3 such that $V_2 - V_3 = \Delta\phi > 0$. Which of the following statements is/are correct about the electric field strengths between the plates and the charges on them? (Assume the plates were initially uncharged)

Professor PC Thomas Classes · Answer

Let the potentials of plates 1, 2, 3, and 4 be $V_1, V_2, V_3, V_4$ respectively. Since 1 and 4 are connected, $V_1 = V_4 = 0$ (taking it as the reference zero potential). Given $V_2 - V_3 = \Delta\phi$. Let $V_2 = V$, then $V_3 = V - \Delta\phi$. The plates 1 and 4 form an isolated system, so the net charge on them is zero: $Q_1 + Q_4 = 0$. The charge on plate 1 resides entirely on its inner face and is $Q_1 = A\epsilon_0 \frac{V_1 - V_2}{d} = -A\epsilon_0 \frac{V}{d}$. The charge on plate 4 resides entirely on its inner face and is $Q_4 = A\epsilon_0 \frac{V_4 - V_3}{d} = A\epsilon_0 \frac{0 - (V - \Delta\phi)}{d} = -A\epsilon_0 \frac{V - \Delta\phi}{d}$. Thus, $-A\epsilon_0 \frac{V}{d} - A\epsilon_0 \frac{V - \Delta\phi}{d} = 0 \implies 2V = \Delta\phi \implies V = \frac{\Delta\phi}{2}$. Therefore, $V_2 = \frac{\Delta\phi}{2}$ and $V_3 = -\frac{\Delta\phi}{2}$. The electric fields (magnitudes) are: $E_{12} = \frac{|V_1 - V_2|}{d} = \frac{\Delta\phi}{2d}$ (A is correct); $E_{23} = \frac{|V_2 - V_3|}{d} = \frac{\Delta\phi}{d}$ (B is correct). The total charge per unit area on plate 2 is $\sigma_2 = \sigma_{2L} + \sigma_{2R} = \epsilon_0 E_{12} + \epsilon_0 E_{23} = \epsilon_0(\frac{\Delta\phi}{2d}) + \epsilon_0(\frac{\Delta\phi}{d}) = \frac{3\epsilon_0 \Delta\phi}{2d}$ (C is correct). The charge per unit area on the inner face of plate 1 is $\sigma_{1R} = \frac{Q_1}{A} = -\epsilon_0 \frac{V}{d} = -\frac{\epsilon_0 \Delta\phi}{2d}$ (D is correct).