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Two plates of the parallel plate capacitor shown below are 6 mm apart and have an area of 5 x 10^-2 m^2 - NSC Technical Sciences - Question 8 - 2022 - Paper 1
Question 8
Two plates of the parallel plate capacitor shown below are 6 mm apart and have an area of 5 x 10^-2 m^2. A potential difference of 100 V is applied across the plates... show full transcript
Worked Solution & Example Answer:Two plates of the parallel plate capacitor shown below are 6 mm apart and have an area of 5 x 10^-2 m^2 - NSC Technical Sciences - Question 8 - 2022 - Paper 1
Step 1
Calculate the charge on EACH plate
Answer
To calculate the charge on each plate of the capacitor, we first need to determine the capacitance using the formula:
Where:
- (\epsilon = 8.85 \times 10^{-12} \text{ F/m} ) (permittivity of free space)
- (A = 5 \times 10^{-2} \text{ m}^2)
- (d = 6 \text{ mm} = 6 \times 10^{-3} \text{ m})
Substituting the values into the equation:
Now, we can find the charge (Q) using the relation:\n[Q = C \times V = 7.38 \times 10^{-11} \times 100 = 7.38 \times 10^{-9} \text{ C}]
Step 2
How will this change affect the magnitude of the capacitance?
Answer
When the distance between the plates is doubled, the capacitance is affected by the relationship:
This indicates that if (d) is increased by a factor of 2, the capacitance will decrease by half. Therefore, we can express this ratio as:
Step 3
Explain the answer to QUESTION 8.2.1
Answer
The decrease in capacitance due to the increase in distance between the plates indicates that capacitance is inversely proportional to the distance. As the plates are moved further apart, the ability of the capacitor to hold charge is reduced, making it necessary to apply a greater potential difference to achieve the same charge, thus confirming our earlier conclusion.