Answer: a. 6.52*EXP{-10}C/m. b. 2.28*EXP{-10}C. c. 38picoFarad.
d. 6.84*EXP{-10}joules
Explanation: We first calculate for the capacitance first. For a concetric cylinder with two radius R1 and R2 the capacitance C is given as
C= {2*pi*permitivity of
freespace*lenght}/In(R2/R1) from the given Question R2 is 2mm and R1 is 1.2mm, lenght is 0.35meter, permitivity of free space is 8.85*EXP {-12} and pi is 3.142.
Therefore Capacitance would be,
C = 2*3.142*8.85*EXP
{-12}*0.35/In(2/1.2)
C = 3.8*EXP {-11} Farad which is also
38*EXP {-12} Farad or 38picoFarad.
Next, we solve for our total charge Q. Charge, capacitance and voltage are related by
Q = C*V = 38*EXP {-12}*6
=2.28*EXP {-10} Coulombs
Next, we obtian charge per unit lenght, which is
Q/L = 2.28*EXP {-10}/0.35
= 6.52*EXP {-10} Coulombs/meter
Next, we obtain the energy stored in the capacitor from
Energy stored = 1/2*(C*V²)
=1/2*38*EXP {-12}*6²
=6.84*EXP {-10} Joules
Note: EXP means 10^
