Please show me how to do this and work. Thanks. A horizontal, parallel-plate cap

Question

Please show me how to do this and work. Thanks.
A horizontal, parallel-plate capacitor with vacuum between itsplates has a capacitance of 26.0 µF.A nonconducting liquid with dielectric constant 4.7 is poured into the space between the plates,filling up a fraction f of its volume. (d) Charges of magnitude 300C are placed on the plates of the partially filledcapacitor. What can you determine about the induced charge on thefree upper surface of the liquid? How does this charge depend onf? Select all that apply to the induced chargeQinduced. 6 Qinduced = 288.46µF
inversely porportional tof
same insign as charge on bottom plate
independent of f
same insign as charge on top plate
Qinduced = 236.17µF
Qinduced = 300µF
directlyproportional to f
6 Qinduced = 288.46µF
inversely porportional tof
same insign as charge on bottom plate
independent of f
same insign as charge on top plate
Qinduced = 236.17µF
Qinduced = 300µF
directlyproportional to f
6

Explanation / Answer

(a)capacitance of parallel plate capacitor C = (oA/d) A = (V/d) or C = (oV/d^2) --------------(1) when a fraction f of its volume is filled with nonconductingliquid then C1 = (koA/d) A = (V1/f * d) or C1 = (koV1/f * d^2) (b)when f = 0 the capacitance is C1 = (koV1/0 * d^2) =(koV1/0) = (c)when f = 1 the capacitance is C1 = (koV1/f * d^2) =(koV1/1 * d^2) = (koV1/d^2) from (1) (o/d^2) = (C/V) or C1 = (C/V) * (V1/f) V1 = f * V or C1 = (C/V) * (f * V/f) = C yes,the expression from part (a) agreewith the answer. (d)the charge density,(Qo/A),where Qo is the total chargeproduced on the surface area A of each plate,is directlyproportional to the electric field. (Qo/A) = -oE = o*(V/d) or Qo = (Ao/d) * V = Co * V o= 8.85 * 10^-12 C^2/Nm^2,Co= 26.0 F =26.0 * 10^-6 F and V is the potential difference between the platesof the capacitor. Qo is independent of f because the initial capacitance Co doesnot depend on f. the induced charge is same in sign as charge on bottomplate the induced charge is independent of "f" the induced charge is same in sign as charge on topplate the induced charge is same in sign as charge on topplate

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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