The figure below shows a parallel plate capacitor of plate area A = 130 cm2 and

Question

The figure below shows a parallel plate capacitor of plate area A = 130 cm2 and plate separation d = 1.40 cm. A potential difference of V0 = 86.0 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 = 3.00 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

part A:

Capacitance C = eoA K /(Kd - b(K-1)

C = (8.85*10^-12 * 130 *10^-4 * 3/((3* 0.014) - 0.078 *(3-1))

C = 3.02 * 10^-12 F
-----------------------

Charge Q = CV

Q = 3.02 *10^-12 * 86

Q = 0.26 nC

-------------------------------

E = Q/eo A

E = 0.26 *10^-9/(8.85 *10^-12* 130 *10^-4)

E = 2.259*10^3 N?C

----------------------------
E remains unchanged and is = 2.259 *10^3 N/C

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