The figure below shows a parallel plate capacitor of plate area A = 100 cm 2 and

Question

The figure below shows a parallel plate capacitor of plate area A = 100 cm2 and plate separation d = 1.10 cm. A potential difference of V0 = 60.0 V is applied between the plates. Suppose that the battery remains connected while the dielectric slab of thickness b = 0.780 cm and dielectric constant = 2.70 is being introduced. Calculate the following values.

(a) the capacitance
pF

(b) the charge on the capacitor plates
nC

(c) the electric field in the gap
N/C

(d) the electric field in the slab, after the slab is in place
N/C

Explanation / Answer

a)   initial capacitance = 8.854 * 10-12 * 0.01/0.0110

                                     =   8.05 * 10-12 F    =   8.05 pF

=> capacitance with dielectric slab = C1 * C2/(C1 + C2)

                                                         = 30.648 * 27.668/(30.648 + 27.668)

                                                          = 14.541 pF

b)   Initial charge on the capacitor plates   =   8.05 * 60.0 = 483 pC = 0.483 nC

   Final charge   on the capacitor plates   =  14.541 * 60.0 = 872.46 pC = 0.87246 nC

c) the electric field in the gap = 0.87246 * 10-9/(10-2 * 8.854 * 10-12)

                                                    =   9853.85 N/C

d)   the electric field in the slab, after the slab is in place = 9853.85/2.70   = 3649.57 N/C

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.