Question

The rechargeable batteries needed in laptop computers are both heavy and expensive. Moreover, the lifetime of many rechargeable batteries is relatively short—perhaps at best a few thousand charge-discharge cycles. Because of this, capacitors have been considered as one possible option to store the required energy. Capacitors are quickly charged, and because no chemical reaction is involved, they have a much longer lifetime than do batteries. Let us say your ultra-low-power laptop requires a potential difference of 8 V to run, and you’d like it to run for a minimum of 4 h using an average power of 1 W. Assume it is possible to “tap” a constant 8 V from a capacitor that has a potential difference greater than or equal to 8 V, but whose potential difference cannot exceed 48 V.

(a) In order to use this capacitor in your laptop, what would its capacitance have to be?

(b) If this were a parallel-plate capacitor, how large would the plate area have to be if the plates were separated by a Mylar layer 0.05 mm thick? Does this type of capacitor seem a feasible alternative to today’s rechargeable batteries?

Answer 1

laptop requires a potential difference of 8 V to run, and you’d like it to run for a minimum of 4 h

=(4*3600s=14,400s) using an average power of 1 W.

current needed by Laptop =i = power / voltage

= 1/8 =0.125 A

Charge needed for 4 h=q = i*t =0.125*14,400 =1,800 C

1. Capacitance needed of capacitor in your laptop= C

= q/V = 1,800/ 8 = 225 F

2. If this were a parallel-plate capacitor and plates were separated by a Mylar layer 0.05 mm thick

d=0.05 mm= 5 x 10^-5m

dielectric constant of Mylar layer=k = 3.2

plate area = A= C *d / K* ϵo

= 225 *(5 x 10^-5) / (3.2)* (8.85 x 10^-12)

= 3.9 x 10^7 m^2

this type of capacitor with this much large plate area does not seem a feasible alternative to today’s rechargeable batteries.