Question

12. In a parallel plate capacitor with air
between the plates, each plate has an
area of 6 x 10-3 m² and the separation
between the plates is 2 mm. a) Calculate
the capacitance of the capacitor, b) If this
capacitor is connected to 100 V supply,
what would be the charge on each plate?
c) How would charge on the plates be
affected if a 2 mm thick mica sheet of
k=6 is inserted between the plates while
the voltage supply remains connected ?​

Answer 1

Answer:

Area of each plate A=6×10−3m2

Distance between the plates d=3mm=0.003m

Capacitance of capacitor C=dAϵo

∴ C=0.0036×10−3×8.85×10−12=17.7pF

Voltage across the capacitor V=100 volts

Thus charge on each plate Q=CV

∴ Q=17.7×10−12×100

⟹ Q=1.77×10−9 C

c) When the voltage supply connected. The capacitance of the capacitor will become k times

∴C′=kC=6×17.7=106.2pF

the potential difference =100V.

∴ Charge on the capacitor

q′=C′V=106.2×10−12×100=106.2×10−10q

make as brilliant.