Question

Two parallel plates with dielectric placed between the plates are as shown in figure. The resultant capacity of capacitor will be [$A = \text{area of plate}$, $t_1, t_2$ and $t_3$ are thickness of dielectric slabs, $k_1, k_2$ and $k_3$ are dielectric constants.]

• A. $A\varepsilon_0 \left[ \frac{t_1 + t_2 + t_3}{k_1 + k_2 + k_3} \right]$
• B. $A\varepsilon_0 \frac{(k_1 k_2 k_3)}{t_1 t_2 t_3}$
• C. $A\varepsilon_0 \left[ \frac{k_1}{t_1} + \frac{k_2}{t_2} + \frac{k_3}{t_3} \right]$
• D. $\frac{A\varepsilon_0}{\left[ \frac{t_1}{k_1} + \frac{t_2}{k_2} + \frac{t_3}{k_3} \right]}$

Hint:

Always verify whether the dielectric boundaries run parallel or perpendicular to the capacitor plates. Slicing the gap thickness creates a series network (where you sum the inverse values $\frac{t}{k}$), whereas slicing across the face area creates a parallel network (where you sum the direct expressions $\frac{k}{t}$).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The screen image shows a parallel-plate system where the space between the plates is shared by three distinct dielectric slabs. Looking closely at the boundary lines in the figure, the slabs are divided vertically, which means the total cross-sectional area of the plates is sliced into sections, while each slab spans the entire plate separation distance from top to bottom. This arrangement forms a

parallel combination of three separate capacitors. We need to identify the correct mathematical equation for the total equivalent capacitance ($C_{\text{eq}}$).

Step 2: Key Formula or Approach:
For a parallel network of three capacitors, the total equivalent capacitance is found by simply summing the individual capacities together:
$$C_{\text{eq}} = C_1 + C_2 + C_3$$ The standard capacitance formula for a dielectric-filled capacitor is:
$$C = \frac{k\varepsilon_0 A'}{t'}$$ where $k$ is the dielectric constant, $A'$ is the partial plate area covered by that specific dielectric, and $t'$ is the thickness of the slab. According to the structural layout, the individual slabs each have a thickness equal to the full plate separation ($t' = t_1 = t_2 = t_3$), and the area factor is factored out.

Step 3: Detailed Explanation:
Let's express the individual capacitance values for each of the three adjacent dielectric blocks:
$$C_1 = \frac{k_1\varepsilon_0 A}{t_1}, \quad C_2 = \frac{k_2\varepsilon_0 A}{t_2}, \quad C_3 = \frac{k_3\varepsilon_0 A}{t_3}$$ Substitute these expressions directly into the linear parallel combination rule:
$$C_{\text{eq}} = \frac{k_1\varepsilon_0 A}{t_1} + \frac{k_2\varepsilon_0 A}{t_2} + \frac{k_3\varepsilon_0 A}{t_3}$$ Factor out the common constants $\varepsilon_0$ and $A$ from all three terms to match the required format:
$$C_{\text{eq}} = A\varepsilon_0 \left[ \frac{k_1}{t_1} + \frac{k_2}{t_2} + \frac{k_3}{t_3} \right]$$ This derived equation aligns perfectly with the expression shown in option (C).

Step 4: Final Answer:
The resultant capacity of the capacitor is $A\varepsilon_0 \left[ \frac{k_1}{t_1} + \frac{k_2}{t_2} + \frac{k_3}{t_3} \right]$, which corresponds to option (C).