There is another useful system of units, besides the SI/MKS. A system called the CGS (Centimeter-Gram-Second) system. In this system, Coulomb's law is given by F=Qqr2ˆr.F=Qqr2ˆr.
Where the distance r is measured in cm (=10-2 m), F in dynes (=10-5 N) and the charges in electrostatic units (es units), where 1 es units of charge=1[3]×10-9 C=1[3]×10−9 C. The number [3] actually arises from the speed of light in the vacuum which is now taken to be exactly given by c=2.99792458×108 m/sc=2.99792458×108 m/s. An approximate value of c then is c=3×108 m/sc=3×108 m/s.
(i) Show that Coulomb's law in CGS units yields 1 esu of charge=1(dyne)1/2 cm.=1(dyne)1/2 cm. Obtain the dimensions of units of charge in terms of mass M, length L and time T. Show that it is given in terms of fractional powers of M and L.
(ii) Write 1 esu of charge = x C, where x is a dimensionless number. Show that this gives 14πε0=10-9x2Nm2C214πε0=10−9x2Nm2C2 with x=1[3]×10-9x=1[3]×10−9, we have 14πε0=[3]2×109 Nm2C2, 14πε0=(2.99792458)2×109Nm2C2 (exactly)14πε0=[3]2×109 Nm2C2, 14πε0=(2.99792458)2×109Nm2C2 (exactly)
Hint: Use Coulomb's law.
(i) Step 1: Find the dimensions of esu.
From the relation, F=Qqr2=1 dyne=[1 esu of charge]2[1 cm]2F=Qqr2=1 dyne=[1 esu of charge]2[1 cm]2
So, 1 esu of charge=(1 dyne)1/2×1 cm=F1/2.L=[MLT-2]1/2L=(1 dyne)1/2×1 cm=F1/2.L=[MLT−2]1/2L
⇒1 esu of charge=M1/2L3/2T-1⇒1 esu of charge=M1/2L3/2T−1
Thus, esu of charge is represented in terms of fractional powers 1212 of M and 3232 of L.
(ii) Step 2: Put the value of charges in esu.
Let 1 esu of charge = x C, where x is a dimensionless number. Coulomb force on two charges, each of magnitude 1 esu separated by 1 cm is dyne=10-5 N=10−5 N. This situation is equivalent to two charges of magnitude xC separated by 10-2 m.10−2 m.
∴F=14πε0x2(10-2)2=1 dyne=10-5N
∴14πε0=10-9x2Nm2C2
Taking, x=1|3|×109, we get,
14πε0=10-2×|3|2×1010 Nm2C2=9×108 Nm2C2
If |3|→2.99792458, we get, 14πε0=8.98755×108 Nm2C-2.
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