CAPACITOR question with solution (Difficulty level : easy one)

Question-1:A Parallel Plate capacitor has following dimensions
Distance between the plates=10 cm
Area of Plate=2 m2
Charge on each plate=8.85 x 10-10 C
Calculate following
  1. Electric Field outside the plates
  2. Electric Field Between the plates
  3. Capacitance of the capacitor
  4. Energy stored in the capacitor
ε0=8.854 x 10-12 C2N-1m-2

As we know that Electric field outside the plates are zero
Electric field Inside the plates is
E=Q/ε0A =50NC-1

Capacitance=ε0A/d =8.854 x 10-12  x 2/.1=17.6 x 10-11 F

Energy stored in capacitor=(1/2)Q2/C=.5 x 8.85 x 10-10 x 8.85 x 10-10/17.6 x 10-11
=22.125 x 10-10 J

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