The figure below shows a parallel plate capacitor of plate area A = 140 cm2 and

Question

The figure below shows a parallel plate capacitor of plate area A = 140 cm2 and plate separation d = 1.15 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 ? = 3.10 is being introduced. Calculate the following values. http://www.webassign.net/hrw/hrw7_25-18.gif (a) the capacitance Incorrect: Your answer is incorrect. 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

C= K*Eo*A/D, where Eo= 8.854x10-12

where:

K is the dielectric constant of the material,
A is the overlapping surface area of the plates,
d is the distance between the plates, and
C is capacitance

C= 3.1*8.8*10^-12 * 0.014/0.0115

=3.32*10^-11

Charge on plates = Q = CV

=3.32*10^-11 * 60

=1.99*10^-9

V=Ed

E=V/d

=60/0.0115

=5217.39 N/C

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