Multiple choice

The separation between plates is reduced by half and the space between them is of dielectric constant 5 and capacitance 8 pF. Calculate the value of capacitance of parallel plate capacitor when reduced by half.

  1. 1.25 x 10-12 F

  2. 80 pF

  3. 0.8 pF

  4. 8 pF

  5. 0.8 x 1012 F
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Capacitance of parallel plate capacitor with air between the plates is C0 = Є0A/d.  When the separation between the plates is reduced to half, C1 = Є0A/(d / 2) = 2Є0A/d.  Thus, final capacitance is C2 = 10 x 8 pF = 80 pF.

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