Question

net charge on the rod is +3.72 x 10-1 c that on the shell is O2- 2.11P What are (c) E and (d) the direction at r-5.098,7 What is the charge on the (e) interior and () exterior surface of the shel (a) Numbe Cb) (e) Number (e B Units Version 4.24.13.

Answer 1

given

r = 2.11 R₂

R₂ = 10.41 R₁

R₁ = 1.2 mm

R₂ = 10.41 x 1.2 = 12.492

r = 2.11 x 12.492

r = 26.358 mm

r = 0.026358 m

q = Q₁ + Q₂

Q₁ = 3.72 x 10^-12 C

Q₂ = - 2.34 Q₁

Q₂ = - 2.34 x 3.72 x 10^-12 = - 8.7 x 10^-12 C

q = 3.72 x 10^-12 C - 8.7 x 10^-12 C

q = - 4.98 x 10^-12 C

L = 12.3 m

using equation = E

E = q / ( 2 x 3.14 x 8.85 x 10^-12 x 12.3 x 0.026358 )

E = - 4.98 x 10^-12 / ( 2 x 3.14 x 8.85 x 10^-12 x 12.3 x 0.026358 )

E = - 0.2763 N/C

| E | = 0.2763 N/C

b )

the negative sign indicates that field points are inward direction

c )

r = 2.11 R₁

= 2.11 x 1.2

r = 2.532 mm

= 0.002532 m

E = Q₁ / ( 2 x 3.14 x 8.85 x 10^-12 x 12.3 x 0.002532 )

= 3.72 x 10^-12 / ( 2 x 3.14 x 8.85 x 10^-12 x 12.3 x 0.002532 )

E = 2.149 N/C

d )

here sign positive

So electric field is outward direction

e )

Q_in = -Q₁

= - 3.72 x 10^-12 C

Q_in = - 3.72 x 10^-12 C

f )

Q_out = Q₂ - Q_in

= - 8.7 x 10^-12 C + 3.72 x 10^-12 C

Q_out = - 4.98x 10^-12 C