The arrangement appears to be a system of parallel plates connected alternately to terminals

$a$

and

$b$

. Let’s determine the net capacitance.

Key Observations:

1. The plates form a series-parallel combination.
2. The area of each plate is
   $A$ 

   , and the separation between adjacent plates is $d$ 

   .

3. The effective configuration can be reduced to find the equivalent capacitance.

Equivalent Capacitance Derivation:

1. Pairing of Plates:
   • Adjacent plates (connected alternately) act as capacitors.
   • Each capacitor has a capacitance
     $C = \frac{\varepsilon_0 A}{d}$ 

     .

2. Parallel and Series Combination:
   • There are three capacitors in the arrangement, effectively forming a single network.
   • The middle plate shares equal charge with both sides, simplifying to an equivalent capacitance of
     $\frac{\varepsilon_0 A}{d}$ 

     .

Thus, the net capacitance is:

$$C_{\text{net}} = \frac{\varepsilon_0 A}{d}.$$

The charging of a capacitor through a resistor is described by the following equation:

$$Q(t) = Q_{\text{max}} \left( 1 - e^{-t/\tau} \right)$$

Where:

• 
  $Q(t)$ 

  is the charge on the capacitor at time $t$ 

  ,

• 
  $Q_{\text{max}}$ 

  is the maximum (steady-state) charge the capacitor can hold,

• 
  $\tau$ 

  is the time constant, $\tau = R \cdot C$ 

  , where $R$ 

  is the resistance and $C$ 

  is the capacitance,

• 
  $t$ 

  is the time.

Step 1: Given condition (10% of steady charge)

We are told that at time

$t$

, the capacitor has collected 10% of the steady charge, so:

$$Q(t) = 0.1 \cdot Q_{\text{max}}$$

Step 2: Substitute into the charging equation

Substitute

$Q(t) = 0.1 \cdot Q_{\text{max}}$

into the charging formula:

$$0.1 \cdot Q_{\text{max}} = Q_{\text{max}} \left( 1 - e^{-t/\tau} \right)$$

Cancel

$Q_{\text{max}}$

from both sides:

$$0.1 = 1 - e^{-t/\tau}$$

Step 3: Solve for $t$

Rearrange the equation to solve for

$e^{-t/\tau}$

:

$$e^{-t/\tau} = 1 - 0.1 = 0.9$$

Take the natural logarithm of both sides:

$$-\frac{t}{\tau} = \ln(0.9)$$

$$t = -\tau \ln(0.9)$$

Using the fact that

$\ln(0.9) = -\ln(10/9)$

:

$$t = \tau \ln\left(\frac{10}{9}\right)$$

Final Answer:

The time at which the capacitor has collected 10% of the steady charge is

$t = \tau \ln\left(\frac{10}{9}\right)$

.

The initial capacitance is \(C_0 = \frac{\varepsilon_0 A}{d}\). With a dielectric slab of thickness \(t = d/4\), the new capacitance is \(C = frac{\varepsilon_0 A}{d - t + \frac{t}{\varepsilon_r}} = \frac{\varepsilon_0 A}{\frac{3d}{4} + \frac{d}{4\varepsilon_r}}\). Simplifying this gives \(C = C_0 \frac{4\varepsilon_r}{3\varepsilon_r + 1}\).

Using \(V = V_0(1 - e^{-t/RC})\), we get \(4 = 12(1 - e^{-t/RC})⇒ e^{-t/RC} = 2/3\). Taking the natural logarithm, \(\frac{t}{RC} = \ln(1.5) = 0.4\), which yields \(C = \frac{10^{-6}}{10 \times 0.4} = 0.25~\mu\text{F}\).

The charging voltage is \( V = V_0(1 - e^{-t/RC})\), so \(120 = 200(1 - e^{-t/RC})\) ⇒ \(e^{t/RC} = 2.5\). This gives \(t/RC = \ln 2.5 = 2.303 \log_{10} 2.5 \approx 0.921\). Solving with \(t = 5~text{s}\) and \(C = 2~\mu\text{F}\) gives \(R \approx 2.7 \times 10^6~\Omega\).

For each capacitor, calculate the maximum charge it can hold: \(Q_1 = C_1 V_1 = (2 \mu\text{F})(3\text{V}) = 6 \mu\text{C}\), \(Q_2 = C_2 V_2 = (3 \mu\text{F})(2\text{V}) = 6 \mu\text{C}\), \(Q_3 = C_3 V_3 = (5 \mu\text{F})(1text{V}) = 5 \mu\text{C}\). In a series combination, the charge must be the same on each capacitor, so the maximum charge is the minimum of these, \(Q_{max} = 5 \mu\text{C}\). The equivalent capacitance in series is \(\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} = \frac{15+10+6}{30} = \frac{31}{30} \text{F}^{-1}\), so \(C_{eq} = \frac{30}{31} \mu\text{F}\). The maximum voltage is \(V_{max} = \frac{Q_{max}}{C_{eq}} = \frac{5 \mu\text{C}}{(30/31) \mu\text{F}} = \frac{5 \times 31}{30} = \frac{31}{6} \text{ Volts}\).