Q62PE - College Physics (Urone)

Question

(a) An $\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\mu F}$capacitor is connected in parallel to another capacitor, producing a total capacitance of $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\mu F}$. What is the capacitance of the second capacitor? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Step-by-Step Solution

Answer

(a) The capacitance of the second capacitor is $-3.00 \mathrm{\mu F}$. 

(b) The other capacitor's capacitance is negative, which is technically impossible.

(c) It is wrong to assume that the capacitors are connected in series.

Step 1: Given data

Total capacitance is $8.00 \mathrm{\mu F}$

Capacitance of one capacitor is $5.00 \mathrm{\mu F}$

Step 2: Capacitors in Parallel

Capacitors in Parallel: The equivalent capacitance C_eq equals the sum of the individual capacitances when capacitors with capacitances C₁, C₂ ... are connected in parallel: 

${\mathbf{C}}_{\mathbf{eq}}\mathbf{=}{\mathbf{C}}_{\mathbf{1}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\left(1\right)$

Step 3: Finding the capacitance of the second capacitor

(a) Equation (1) is used, to find the equivalent capacitance of C₁ and C₂, as they are connected in parallel:

${\mathbf{C}}_{\mathbf{eq}}\mathbf{=}{\mathbf{C}}_{\mathbf{1}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}$

When we solve for C₂, we get: 

${\mathbf{C}}_{\mathbf{2}}\mathbf{=}{\mathbf{C}}_{\mathbf{eq}}\mathbf{-}{\mathbf{C}}_{\mathbf{1}}$

Now, substitute the values for C_eq and C₁ we get:

  $\begin{array}{rcl}{\mathbf{C}}_{\mathbf{2}}& \mathbf{=}& \mathbf{5}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\mu F}\mathbf{-}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\mu F}\\ & \mathbf{=}& \mathbf{-}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\mu F}\end{array}$

                                                           

Therefore, the capacitance of the second capacitor is $-3.00 \mathrm{\mu F}$.

Step 4: Solving for Part (b)

(b) The other capacitor's capacitance is negative, which is physically impossible.

Step 5: Solving for Part (c)

(c) The assumption that the capacitors are linked in parallel is incorrect since the equivalent capacitance of two capacitors connected in parallel is greater than the total of their individual capacitances.