Secure https://session.masteringphysics.com/myct/itemview.. a CChapters 20, 21,

Question

Secure https://session.masteringphysics.com/myct/itemview.. a CChapters 20, 21, 22 Problem Quiz Problem 21.37 C15 of 20 Part C Constants ?at n te electric field before the Tetkn sremoved? Two 2.8 Cm-diameter electrodes with a 0.15-mm thick sheet of Teflon between them are attached to a 9.0 V batery Without disconnecting the batery the Teflon is emoved V/m Submit - Part D What is the charge ater the Tefion is removed? 245 Submit Incorrect: One attempt remaining; Try Again Part E What s the potential difference ather the Tefon s remaved?

Explanation / Answer

Given that
Diameter,
d = 2.8cm
= 0.028m

radius of electrodes ,
r = d/2 = 0.028m/2 = 1.4* 10^-2m

thickness of teflon t = 0.15* 10^-3 m

voltage across the two plates v0 = 9.0v
dielectric constant of teflon k = 2.1
A) charge before teflon is removed
Q =k ?0Av0/t
= (2.1)8.85 * 10^-12 C^2/N.m^2 ( 3.14) ( 1.4 * 10^-2m)^2 * 9.0v/ 0.15 * 10^-3m
= 686.3 * 10^-12 C
--------------_-
B) potential difference before teflon is removed is same because battery is not disconted
V = 9.0v
-------------------
C) intial electric field
V = E0t/k
E0 = Vk/t
= (2.1)9.0v/0.15 * 10^-3m = 126000 V/m
electric field before teflon is removed E = E0/k = 126000V/m/2.1 = 60000V/m
-----------
D)
charge after teflon is removed
Q= Cv0
= ?0Av0/t
=8.85 * 10^-12 C^2/N.m^2 ( 3.14 ( 1.4 * 10^-2m)^2 * 9.0v/ 0.15* 10^-3m
= 326.81 * 10^-12C
------------
E)potential difference afterteflon is removed is same because battery is not disconted
--------------
F) electric field E = V/t
= 9.0v/0.15 * 10^-3m
=60000V/m

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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