Consider the following capacitors: C1 has an area of 4.0 cm2 and a separation di

Question

Consider the following capacitors: C1 has an area of 4.0 cm2
and a separation distance of 2.0 mm;
C2 has an area of 5.0 cm2
and a separation distance of 3.0 mm; C3 has an area of 4.0 cm2
and a
separation distance of 1.0 mm; C4 has an area of 2.0 cm2
and a separation distance of 1.0 mm.
Rank the capacitors based on their capacitance from highest to lowest. Show one sample
calculation of how you determined the capacitance

Each of the capacitors in the previous problem is given a charge of 8.0 µC. Rank the potential of each capacitor from highest to lowest. Show one sample calculation.

Consider the following capacitors: C1 has an area of 4.0 cm2 and a separation distance of 2.0 mm; C2 has an area of 5.0 cm and a separation distance of 3.0 mm; C3 has an area of 4.0 cm2 anda separation distance of1.0mm; C4 has an area of2.0 cm2 and a separation distance of1.0mm. Rank the capacitors based on their capacitance from highest to lowest. Show one sample calculation of how you determined the capacitance. a. b. Each of the capacitors in the previous problem is given a charge of 8.0 uC. Rank the potential of each capacitor from highest to lowest. Show one sample calculation

Explanation / Answer

a)

Capacitance is given by

C=eoA/d

therefore

C1=(8.8542*10-12)(4*10-4)/(2*10-3)=1.77 pF

C2=(8.8542*10-12)(5*10-4)/(3*10-3)=1.48 pF

C3=(8.8542*10-12)(4*10-4)/(1*10-3)=3.54 pF

C4=(8.8542*10-12)(2*10-4)/(1*10-3)=1.77 pF

Therfore

C3>C1=C4>C2

b)

electric potential across each capacitor is

V=Q/C

therefore

V1=(8*10-6)/(1.77*10-12)=4.52*106 V

V2=(8*10-6)/(1.48*10-12)=5.42*106 V

V3=(8*10-6)/(3.54*10-12)=2.26*106 V

V4=(8*10-6)/(1.77*10-12)=4.52*106 V

therefore

V2>V1=V4>V3

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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