2) Two large, parallel conducting plates of area A are separated by a distance d

Question

2) Two large, parallel conducting plates of area A are separated by a distance d. A slab of dielectric material (dielectric constant ?R) with the same area A and a thickness d/3 is inserted symmetrically between the plates, with the center of the slab at d/2 from each plate.
a. What is the capacitance of this system with the slab in place?
b. If a charge Q and –Q is placed on the conducting plates, what is the bound surface charge on the slab?
c. What is the capacitance of the system if the slab is placed right next to one of the plates?
d. What is the bound surface charge on the slab in this case, if Q and –Q are place on the conducting plates?
e. If 2Q is placed on the top plate and –Q is placed on the bottom, what is the bound surface charge on the slab?

Explanation / Answer

the capacitance of the system can be all added togather

let the dielectric constant be k

hence the effective capacitance is 3oA/d[k+2]=C

b)the bound surface charge on the dielectric is Q/C=POTENTIAL DIFFERENCE

electric field is V*d

electric field inside dielectric is vd*k=Qdk/C=surface charge on the dielectric/Ao

hence we have the surface charge

c)is the same

d)in this case the 2Q chargew on the top plate splits into Q on the inner surface and Q on outer and it doesnt matter same answer

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

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