Two long, charged, thin-walled, concentric cylinders have radii of 3.0 and 6.0 cm. The charge per unit length is 6.00 ✕ 10−6 C/m on the inner shell and −7.50 ✕ 10−6 C/m on the outer shell. (a) Find the magnitude and direction of the electric field at radial distance r = 4.7 cm from the common central axis. (Take radially outward to be positive.) N/C (b) Find the magnitude and direction of the electric field at r = 8.1 cm, using the same sign convention. N/C

a) E = 2.30×10^6N/C

Direction =+E . It will point radically outward

b)E = -3.3×10^5NC^-1

Direction =- E,it will point radically inwards

Explanation:

Magnitude E at 4.7cm= 0.047m

E= wavelength 1/2×3.142×rEo

E= (6.0×10^-6)/2×3.142×0.047×(8.85×10^-12)

E= (6.0×10^-6)/(2.6×10^-12)

E= 2.30×10^6N/C

Direction= +E shows it will point radically outward.

b) Magnitude E at 8.1cm=0.081

E= (wavelength 1+ wavelength 2) / 2×3.142rEo

E= [(6.0×10^-6)-(7.5×10^-6)]/(2×3.142×0.081×(8.84×10^-12)

E= -(1.5×10^-6)/(4.5×10^-11)

E= -3.3×10^5N/C

Direction =I-A it will point inwards

prolly a beacuse the momentuem the car has increases while going down hill