Class Management I Help Homework Assignment #20 Begin Date: 3 /25/2018 4:00 00 P

Question

Class Management I Help Homework Assignment #20 Begin Date: 3 /25/2018 4:00 00 PM Due Date 00 PM End Date: 1 (33%) Problem 3: An ideal parallel plate capacitor has cross-sectional area A- 2.8 em2 and separation d 5 mm Ose of the plates has a net electric chage of +55 n/C and the other plate has a net electric charge of-55 nC Two dielectrics, with dielectric 2d constarts K1 _ 4 and K2 = 8, are placed in series in-between the plates of the capacitor What sthe potental energy (in m) of this capacio Grade Semmar sinocos tial detailed vien Degrees Radians Teedback dedcts per feebechk 5

Explanation / Answer

(1).

Here, in the question, two dielectrics are inserted in parallel plates

Given- Charge on plates(Q)= 55nC

Distance between plates (2d)= 2*5mm= 10mm or 10*10^-3m

Height of dielectrics (d)= 5mm= 5*10^-3

Dielectric constant K1 and K2= 4 and 8

Area of plate is (A)= 2.8cm^2= 2.8 *10^-4m^2

Now we have to find potential difference between plates

As there are two dielectrics inserted

Capacitance of individual

C1= (E'*A*K1)/d Where E'= Permiability constant= 8.85*10^-12

C2= (E'*A*K2)/d

C=C1*C2/(C1+C2).....eq(1)

Inserting values of C1 and C2 in eq(1)

We will get C=(K1*K2*E'*A)/d(K1+K2)

C= (4*8*8.85*10^-12*2.8*10^-4)/5*10^-3*12

C=13.2*10^-13

Now we have a capacitance of system now we have to find the potential difference(V)

As we know Q=CV

or V=Q/C

V=(5.5*10^-8)/(13.2*10^-13)

V=0.41*10^3= 416.6v

Potential difference is = 416.6volt

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---