A ball of charge containing 0.47 C has radius 0.26 m. a) What is the electric field strength at a radius of 0.18 m? b) What is E at 0.50 m? c) What is E at infinity? d) What is …

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asked Apr 10, 2020 59.3k views

1 Answer

Rsteckly · Answer 1 · 2020-04-15T14:08:45+0000

Answer:

a) 4.325*10^10 V/m

b) 1.689*10^10 V/m

c) 0

d) 6.384 C/m^3

Step-by-step explanation:

Hello!

The electric field of a sphere of uniform charge is a piecewise function, let a be the raius of the sphere

For r<a:

E = kQ(r)/(a^(3))

For r>a:

E=kQ/r^(2)

Since a=0.26m and k= 8.987×10⁹ N·m²/C²

a)

E= 8.987*10^(9)(0.47C *0.18m)/((0.26m)^(3))

E=4.325*10^10 V/m

b)

E= 8.987*10^(9)(0.47C)/((0.5m)^(2))

E=1.689*10^10 V/m

c)

Since r --> ∞ 1/r^2 --> 0

E(∞)=0

d)

The charge density may be obtained dividing the charge by the volume of the sphere:

$\rho = (Q)/(V) =(0.47C)/((4)/(3) \pi (0.26m)^(3))=6.384 C/m^(3)$