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Consider a parallel-plate capacitor with capacitance CC, what happens when we double the separation between the plates and halve the area of each plate?

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Q. Consider a parallel-plate capacitor with capacitance CC, what happens when we double the separation between the plates and halve the area of each plate?
  1. Capacitance Formula: Capacitance of a parallel-plate capacitor is given by C=ε0×(Ad)C = \varepsilon_0 \times \left(\frac{A}{d}\right), where ε0\varepsilon_0 is the permittivity of free space, AA is the area of the plates, and dd is the separation between the plates.
  2. Doubling Separation: If we double the separation, the new separation is 2d2d.
  3. Halving Area: If we halve the area of each plate, the new area is A/2A/2.
  4. Calculating New Capacitance: The new capacitance C′C' is then given by C′=ε0⋅(A/22d)C' = \varepsilon_0 \cdot \left(\frac{A/2}{2d}\right).
  5. Simplifying Expression: Simplify the expression for C′C' to get C′=ε0⋅(A4d)C' = \varepsilon_0 \cdot \left(\frac{A}{4d}\right).
  6. Comparing Capacitance: Comparing the new capacitance C′C' with the original capacitance CC, we have C′=C/4C' = C/4.
  7. Conclusion: Therefore, the new capacitance is 14\frac{1}{4} of the original capacitance.

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