QUARTER FINAL STAGE 2024
YILO KROBO
KOFORIDUA SENIOR HIGH TECH
KNUST SHS
ROUND 1
QUESTION
The capacitance of a capacitor of plate area $1.00\text{ m}^2$ and plate separation $1.00\text{ mm}$ is $8.85\times10^{-9}\text{ F}$ when the space between the plates is filled with air.
Find the capacitance of a capacitor of plate area $A$ and plate separation $D$ that is filled with a fluid of dielectric constant $K$.
FORMULA: $C=\dfrac{K\varepsilon_0A}{D}$, where $\varepsilon_0=8.85\times10^{-12}\text{ F/m}$
1. $A=3.20\text{ m}^2$, $D=2.00\text{ mm}$, $K=1.50$
ANSWER: $C = 2.12\times10^{-8}\text{ F}$
SOLUTION 1:
$d = 2\times10^{-3}\text{ m}$
$C = \dfrac{15\times10^{-1}\times885\times10^{-14}\times32\times10^{-1}}{2\times10^{-3}}$
$C = 15\times885\times16\times10^{-1-14-1+3}$
$C = 212400\times10^{-13}$
$C = 2.12\times10^{-8}\text{ F}$
SOLUTION 2:
$d = 2\times10^{-3}\text{ m}$
$C = \dfrac{15\times10^{-1}\times885\times10^{-14}\times32\times10^{-1}}{2\times10^{-3}}$
$C = 15\times885\times16\times10^{-1-14-1+3}$
$C = 212400\times10^{-13}$
$C = 2.12\times10^{-8}\text{ F}$
2. $A=3.60\text{ m}^2$, $D=3.00\text{ mm}$, $K=1.20$
ANSWER: $C = 1.27\times10^{-8}\text{ F}$
SOLUTION 1:
$d = 3\times10^{-3}\text{ m}$
$C = \dfrac{12\times10^{-1}\times885\times10^{-14}\times36\times10^{-1}}{3\times10^{-3}}$
$C = 12\times885\times12\times10^{-1-14-1+3}$
$C = 127440\times10^{-13}$
$C = 1.27\times10^{-8}\text{ F}$
SOLUTION 2:
$d = 3\times10^{-3}\text{ m}$
$C = \dfrac{12\times10^{-1}\times885\times10^{-14}\times36\times10^{-1}}{3\times10^{-3}}$
$C = 12\times885\times12\times10^{-1-14-1+3}$
$C = 127440\times10^{-13}$
$C = 1.27\times10^{-8}\text{ F}$
3. $A=3.00\text{ m}^2$, $D=1.20\text{ mm}$, $K=2.00$
ANSWER: $C = 4.42\times10^{-8}\text{ F}$
SOLUTION 1:
$d = 12\times10^{-4}\text{ m}$
$C = \dfrac{2\times8850\times10^{-15}\times3}{12\times10^{-4}}$
$C = \dfrac{53100\times10^{-15}}{12\times10^{-4}}$
$C = 4425\times10^{-11}$
$C = 4.42\times10^{-8}\text{ F}$
SOLUTION 2:
$d = 12\times10^{-4}\text{ m}$
$C = \dfrac{2\times8850\times10^{-15}\times3}{12\times10^{-4}}$
$C = \dfrac{53100\times10^{-15}}{12\times10^{-4}}$
$C = 4425\times10^{-11}$
$C = 4.42\times10^{-8}\text{ F}$
NOTE: SOLUTION 2 repeats SOLUTION 1 exactly for all three items.
NOTE: In item 3, $4.425\times10^{-8}$ rounds half-up to $4.43\times10^{-8}\text{ F}$; the stated value is $4.42\times10^{-8}\text{ F}$.
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PRACTICE QUESTIONS
1. QUESTION: Find the capacitance of a parallel-plate capacitor of plate area 0.500 m² and plate separation 0.500 mm when the space between the plates is filled with a fluid of dielectric constant 1.50.
ANSWER: $1.33\times10^{-8}$ F
SOLUTION:
$C = \dfrac{K\varepsilon_0 A}{d}$
$d = 0.500\times10^{-3}\text{ m}$
$C = \dfrac{1.50 \times 8.85\times10^{-12} \times 0.500}{0.500\times10^{-3}}$
$C = \dfrac{15\times10^{-1} \times 5\times10^{-1} \times 8.85\times10^{-12}}{5\times10^{-4}}$
$C = 3 \times 5 \times 8.85\times10^{-12} \times10^{2}$
$C \approx 1.33\times10^{-8}\text{ F}$
2. QUESTION: Find the capacitance of a parallel-plate capacitor of plate area 2.00 m² and plate separation 4.00 mm when the space between the plates is filled with a fluid of dielectric constant 5.00.
ANSWER: $2.21\times10^{-8}$ F
SOLUTION:
$C = \dfrac{K\varepsilon_0 A}{d}$
$d = 4.00\times10^{-3}\text{ m}$
$C = \dfrac{5.00 \times 8.85\times10^{-12} \times 2.00}{4.00\times10^{-3}}$
$C = \dfrac{5 \times 2 \times 8.85\times10^{-12}}{4\times10^{-3}}$
$C = \dfrac{5 \times 8.85\times10^{-12}}{2} \times10^{3}$
$C \approx 2.21\times10^{-8}\text{ F}$
3. QUESTION: Find the capacitance of a parallel-plate capacitor of plate area 1.00 m² and plate separation 1.00 mm when the space between the plates is filled with a fluid of dielectric constant 8.00.
ANSWER: $7.08\times10^{-8}$ F
SOLUTION:
$C = \dfrac{K\varepsilon_0 A}{d}$
$d = 1.00\times10^{-3}\text{ m}$
$C = \dfrac{8.00 \times 8.85\times10^{-12} \times 1.00}{1.00\times10^{-3}}$
$C = \dfrac{8 \times 8.85\times10^{-12}}{10^{-3}}$
$C = 8 \times 8.85\times10^{-12} \times10^{3}$
$C = 7.08\times10^{-8}\text{ F}$
4. QUESTION: Find the capacitance of a parallel-plate capacitor of plate area 3.00 m² and plate separation 3.00 mm when the space between the plates is filled with a fluid of dielectric constant 6.00.
ANSWER: $5.31\times10^{-8}$ F
SOLUTION:
$C = \dfrac{K\varepsilon_0 A}{d}$
$d = 3.00\times10^{-3}\text{ m}$
$C = \dfrac{6.00 \times 8.85\times10^{-12} \times 3.00}{3.00\times10^{-3}}$
$C = \dfrac{6 \times 3 \times 8.85\times10^{-12}}{3\times10^{-3}}$
$C = 2 \times 3 \times 8.85\times10^{-12} \times10^{3}$
$C = 5.31\times10^{-8}\text{ F}$
5. QUESTION: Find the capacitance of a parallel-plate capacitor of plate area 1.20 m² and plate separation 1.20 mm when the space between the plates is filled with a fluid of dielectric constant 1.20.
ANSWER: $1.06\times10^{-8}$ F
SOLUTION:
$C = \dfrac{K\varepsilon_0 A}{d}$
$d = 1.20\times10^{-3}\text{ m}$
$C = \dfrac{1.20 \times 8.85\times10^{-12} \times 1.20}{1.20\times10^{-3}}$
$C = \dfrac{12\times10^{-1} \times 12\times10^{-1} \times 8.85\times10^{-12}}{12\times10^{-4}}$
$C = 12 \times 8.85\times10^{-12} \times10^{2}$
$C \approx 1.06\times10^{-8}\text{ F}$
6. QUESTION: Find the capacitance of a parallel-plate capacitor of plate area 2.50 m² and plate separation 2.50 mm when the space between the plates is filled with a fluid of dielectric constant 2.00.
ANSWER: $1.77\times10^{-8}$ F
SOLUTION:
$C = \dfrac{K\varepsilon_0 A}{d}$
$d = 2.50\times10^{-3}\text{ m}$
$C = \dfrac{2.00 \times 8.85\times10^{-12} \times 2.50}{2.50\times10^{-3}}$
$C = \dfrac{2 \times 25\times10^{-1} \times 8.85\times10^{-12}}{25\times10^{-4}}$
$C = 2 \times 8.85\times10^{-12} \times10^{3}$
$C = 1.77\times10^{-8}\text{ F}$
7. QUESTION: Find the capacitance of a parallel-plate capacitor of plate area 4.00 m² and plate separation 5.00 mm when the space between the plates is filled with a fluid of dielectric constant 4.00.
ANSWER: $2.83\times10^{-8}$ F
SOLUTION:
$C = \dfrac{K\varepsilon_0 A}{d}$
$d = 5.00\times10^{-3}\text{ m}$
$C = \dfrac{4.00 \times 8.85\times10^{-12} \times 4.00}{5.00\times10^{-3}}$
$C = \dfrac{4 \times 4 \times 8.85\times10^{-12}}{5\times10^{-3}}$
$C = \dfrac{4 \times 4 \times 8.85\times10^{-12}}{5} \times10^{3}$
$C \approx 2.83\times10^{-8}\text{ F}$
8. QUESTION: Find the capacitance of a parallel-plate capacitor of plate area 1.50 m² and plate separation 2.00 mm when the space between the plates is filled with a fluid of dielectric constant 3.00.
ANSWER: $1.99\times10^{-8}$ F
SOLUTION:
$C = \dfrac{K\varepsilon_0 A}{d}$
$d = 2.00\times10^{-3}\text{ m}$
$C = \dfrac{3.00 \times 8.85\times10^{-12} \times 1.50}{2.00\times10^{-3}}$
$C = \dfrac{3 \times 15\times10^{-1} \times 8.85\times10^{-12}}{2\times10^{-3}}$
$C = \dfrac{3 \times 15 \times 8.85\times10^{-12}}{2} \times10^{2}$
$C \approx 1.99\times10^{-8}\text{ F}$
9. QUESTION: Find the capacitance of a parallel-plate capacitor of plate area 2.50 m² and plate separation 1.50 mm when the space between the plates is filled with a fluid of dielectric constant 6.00.
ANSWER: $8.85\times10^{-8}$ F
SOLUTION:
$C = \dfrac{K\varepsilon_0 A}{d}$
$d = 1.50\times10^{-3}\text{ m}$
$C = \dfrac{6.00 \times 8.85\times10^{-12} \times 2.50}{1.50\times10^{-3}}$
$C = \dfrac{6 \times 25\times10^{-1} \times 8.85\times10^{-12}}{15\times10^{-4}}$
$C = \dfrac{6 \times 5 \times 8.85\times10^{-12}}{3} \times10^{3}$
$C = 2 \times 5 \times 8.85\times10^{-12} \times10^{3}$
$C = 8.85\times10^{-12} \times10^{4}$
$C = 8.85\times10^{-8}\text{ F}$
10. QUESTION: Find the capacitance of a parallel-plate capacitor of plate area 0.500 m² and plate separation 1.00 mm when the space between the plates is filled with a fluid of dielectric constant 2.50.
ANSWER: $1.11\times10^{-8}$ F
SOLUTION:
$C = \dfrac{K\varepsilon_0 A}{d}$
$d = 1.00\times10^{-3}\text{ m}$
$C = \dfrac{2.50 \times 8.85\times10^{-12} \times 0.500}{1.00\times10^{-3}}$
$C = \dfrac{25\times10^{-1} \times 5\times10^{-1} \times 8.85\times10^{-12}}{10^{-3}}$
$C = 25 \times 5 \times 8.85\times10^{-12} \times10^{1}$
$C \approx 1.11\times10^{-8}\text{ F}$
11. QUESTION: Find the capacitance of a parallel-plate capacitor of plate area 1.20 m² and plate separation 5.00 mm when the space between the plates is filled with a fluid of dielectric constant 8.00.
ANSWER: $1.70\times10^{-8}$ F
SOLUTION:
$C = \dfrac{K\varepsilon_0 A}{d}$
$d = 5.00\times10^{-3}\text{ m}$
$C = \dfrac{8.00 \times 8.85\times10^{-12} \times 1.20}{5.00\times10^{-3}}$
$C = \dfrac{8 \times 12\times10^{-1} \times 8.85\times10^{-12}}{5\times10^{-3}}$
$C = \dfrac{8 \times 12 \times 8.85\times10^{-12}}{5} \times10^{2}$
$C \approx 1.70\times10^{-8}\text{ F}$
12. QUESTION: Find the capacitance of a parallel-plate capacitor of plate area 1.00 m² and plate separation 4.00 mm when the space between the plates is filled with a fluid of dielectric constant 1.50.
ANSWER: $3.32\times10^{-9}$ F
SOLUTION:
$C = \dfrac{K\varepsilon_0 A}{d}$
$d = 4.00\times10^{-3}\text{ m}$
$C = \dfrac{1.50 \times 8.85\times10^{-12} \times 1.00}{4.00\times10^{-3}}$
$C = \dfrac{15\times10^{-1} \times 8.85\times10^{-12}}{4\times10^{-3}}$
$C = \dfrac{15 \times 8.85\times10^{-12}}{4} \times10^{2}$
$C \approx 3.32\times10^{-9}\text{ F}$