A copper slab of thickness b = 1.194 mm is inserted at constant speed into a par

Question

A copper slab of thickness b = 1.194 mm is inserted at constant speed into a parallel plate capacitor with gap d = 8.0 mm, as shown in the figure The slab 15 centered exactly halfway between the plates. If the Initial capacitance was C = 4.00E-11 F, what is the capacitance after the slab is introduced? If a charge q = 6.00E-6 C is maintained on the plates the slab is inserted, what is the ratio of the initial stored energy to the final stored energy? How much work is done by the person Inserting the slab?

Explanation / Answer

(a)
Initial Capacitance, A*e0/d = 4.0 * 10^-11 F

The new arrangement is equivalent to having two capacitors connected in series -
C' = A*e0/((d - b)/2)
Ct = C'/2
Ct = A*e0/((d - b))
Ct = (d * 4.0 * 10^-11 ) / (d-b)
Ct = (8 * 4.0 * 10^-11) / (8 - 1.194)
Ct = 4.70 * 10^-11 F

(b)
Initial Stored Energy = 1/2 * Q^2/C = 1/2 * ((6*10^-6)^2 / (4.0 * 10^-11)) = 0.45 J
Final Stored Energy = 1/2 * Q^2/Ct = 1/2 * ((6*10^-6)^2 / (4.7 * 10^-11)) = 0.383 J

Ratio of Initial Stored Energy/ Final Stored Energy = 0.45/0.383
Ratio of Initial Stored Energy/ Final Stored Energy = 1.175

(c)
Work done  = 0.383 - 0.45 J
Work done = - 0.067 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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