Question:
Capacitance $C$ of a parallel-plate structure is calculated as $20~\text{pF}$ using $C=\dfrac{\varepsilon_0\varepsilon_r A}{d}$. The value is then measured using an ideal LCR meter (no cable-capacitance error). Which reading is likely to be correct?
Capacitance $C$ of a parallel-plate structure is calculated as $20~\text{pF}$ using $C=\dfrac{\varepsilon_0\varepsilon_r A}{d}$. The value is then measured using an ideal LCR meter (no cable-capacitance error). Which reading is likely to be correct?
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Parallel-plate formula neglects edges; fringing always increases $C$ slightly, so a precise measurement is typically just above the calculated value.
Updated On: Sep 1, 2025
- 20.5 pF
- 20 pF
- 19.5 pF
- 10 pF
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The Correct Option is A
Solution and Explanation
Step 1: Ideal formula vs. real capacitor.
$C=\dfrac{\varepsilon_0\varepsilon_r A}{d}$ assumes an {ideal} parallel-plate capacitor with uniform field and {no fringing}.
Step 2: Account for fringing. A real capacitor has edge (fringing) fields that add an extra capacitance $C_f>0$. Hence the actual capacitance is \[ C_{\text{actual}} = C_{\text{ideal}} + C_f>C_{\text{ideal}}. \] Step 3: Choose the reading. With $C_{\text{ideal}}=20~\text{pF}$ and an ideal LCR meter (no cable/parasitic subtraction needed), the measured value should be slightly {higher} than $20~\text{pF}$. The closest option is 20.5 pF.
Final Answer: 20.5 pF
Step 2: Account for fringing. A real capacitor has edge (fringing) fields that add an extra capacitance $C_f>0$. Hence the actual capacitance is \[ C_{\text{actual}} = C_{\text{ideal}} + C_f>C_{\text{ideal}}. \] Step 3: Choose the reading. With $C_{\text{ideal}}=20~\text{pF}$ and an ideal LCR meter (no cable/parasitic subtraction needed), the measured value should be slightly {higher} than $20~\text{pF}$. The closest option is 20.5 pF.
Final Answer: 20.5 pF
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