A parallel plate capacitor has plates of area A = 250 cm^2 and separation d = 2.
ID: 1438560 • Letter: A
Question
A parallel plate capacitor has plates of area A = 250 cm^2 and separation d = 2.00 cm. The capacitor is charged to a potential difference V_o = 120 V. Then the battery is disconnected (the charge Q on the plates won't change), and a dielectric sheet (k = 3.25) of the same area A but thickness l = 5.00 mm is placed between the plates as shown in the figure below Determine: The initial capacitance of the air filled capacitor. The charge on each plate before the dielectric is inserted. The charge induced on each face of the dielectric after it is inserted. The electric field in the space between each plate and the dielectric. The electric field in the dielectric The potential difference between the plates after the dielectric is added The capacitance after the dielectric is in placeExplanation / Answer
a. The initial Capacitance is C= A/d = 250*10-4/2*10-2 = 1.25 F here we assume e0 =1.
b. Charge on each plate Q = C*V = 1.25*120 C = 150 C
c. Induced Surfacw charge on dielectric qi = Q*(1 - k-1) = 103.846 C
d. The electric field in space between plate and dielectric = 120/2*10-2 = 6000N/C
e. The electric field in dielectric = 6000/3.25 N/C = 1846.15 N/C
g. The capacitance after dielectric is in place Cnew = (C1-1 + C2-1 + C3-1)-1 [Since this is an equivalent parallel connection of capacitors] = [(A/{(d-0.5)/2})-1+(kA/0.5)-1 + (A/{(d-0.5)/2})-1]-1 = 0.015 F
f. so potential difference between plates = Q/Cnew = 150/0.015 V = 10,000V