(c26p74) A slab of copper of thickness b = 1.860 mm is thrust into a parallel-pl

Question

(c26p74) A slab of copper of thickness b = 1.860 mm is thrust into a parallel-plate capacitor of C = 1.00×10-11 F of gap d = 8.0 mm, as shown in the figure; it is centered exactly halfway between the plates. If a charge q = 2.00×10-6 C is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted? Tries 0/10 How much work is done on the slab as it is inserted? Tries 0/10 Is the slab pulled in or must it be pushed in? (You have only ONE try for this answer!)

Explanation / Answer

b = 1.860 mm = 1.86 X 10-3 m

C = 1.00 X 10-11 F

d = 8.0 mm = 8 X 10-3 m

Charge q = 2.00 X 10-6 C

the ratio of the stored energy before to that after the slab is inserted is

Ui / Uf = ( 0.5 X q2 / Ci ) / ( 0.5 X q2 / Cf )

Ui / Uf = Ci / Cf

Ui / Uf = d / d – b

Ui / Uf = 8 / 8 – 1.86

Ui / Uf = 1.30

Now calculating the work is done on the slab as it is inserted is

W = dU

= Uf - Ui

= 0.5 X q2 / Cf - 0.5 X q2 / Ci

W = 0.5 X q2 ( 1 / Cf – 8 X 10-3 / 8.85 X 10-12 X 1 X 10-3 )

But here A value not mentioned taken as 1 mm2

And finding separately Cf value which is unknown

Cf = 8.85 X 10-12 X 1 X 10-3 / ( 8 – 1.86 ) X 10-3

Cf = 1.44 X 10-12 F

This value substituting in above equation

W = 0.5 X (2.00 X 10-6)2 ( ( 1 / 1.44 X 10-12 ) – ( 8 X 10-3 / 8.85 X 10-12 X 1 X 10-3 ) )

W = 1 X 10-12 ( ( 6.94 X 1011) – 9 X 1011 )

W = 10-12 X 2.036 X 1011

W = 2.06 X 10-1

W = 0.206 J

and last the slab should be inserted

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

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