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 Incorrect: Your answer is incorrect. N/C (d) the electric field in the slab, after the slab is in place Incorrect: Your answer is incorrect. N/C (See Sample Problem 25-7 in the textbook.)

Explanation / Answer

introduction of dielectric slab reduces the capacitance to C/k ,so net capacitance = C(air) in series with C( dielectric) = epsilon0 *A/d + K* epsilon0 *A/d = 8.845*10^-12*140 *10^-4 /(0.37 *10^-2) + 3.1 *8.845*10^-12*140 *10^-4 /(0.78 *10^-2) = 3.34 *10^-11 +4.92* 10^-11 = 8.26*10 ^-11 C/V a) the capacitance = 8.26*10 ^-11 C b) q= V*C =60 * 8.26*10 ^-11 = 4.956 nC c) E =v/d =60 /1.15*10 ^-2 =5217.3 v/m d) E in dielectric= E /k =1683.0 V/m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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