The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distributed uniformly throughout its volume. A particle that carries an unknown charge qpart is located on the x axis at x=+2R. The magnitude of the electric field due to the sphere-particle combination is zero at x=+R/4 on the xaxis.

Part A

What is qpart in terms of qsphere?
Express your answer in terms of qsphere.

qpart =
nothing

SubmitRequest Answer

Part B

At what other locations on the x axis is the electric field zero?
Express your answer in terms of R.

x =
nothing

Question

contestada

Respuesta :

jorgezarate jorgezarate · Answer 1 · 2019-09-13T06:50:15+02:00

Answer:

[tex]q=49Q/64[/tex]

and

[tex]x =16R/15 [/tex]

Explanation:

See attached figure.

[tex]E_{Q}=[/tex] E due to sphere

[tex]E_{q}=[/tex] E due to particule

[tex]E_{total}=E_{Q}-E_{q}=0[/tex] (1)

according to the law of gauss and superposition Law:

[tex]E_{Q}=E_{1}+E_{2}=E_{2}[/tex] ; electric field due to the small sphere with r1=R/4

[tex]E_{Q}=kq_{2}/(r_{1}^{2})=[/tex]

[tex]q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}[/tex]

then: [tex]E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})[/tex] (2)

on the other hand, for the particule:

[tex]E_{q}=kq/(r_{p}^{2})[/tex]

[tex]r_{p}=2R-R/4=7R/4[/tex] ⇒ [tex]E_{q}=16kq/(49R^{2})[/tex] (3)

We replace (2) y (3) in (1):

[tex]E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})[/tex]

[tex]q=49Q/64[/tex]

--------------------

if R<x<2R AND [tex]E_{total}=E_{Q}-E_{q}=0[/tex]

[tex]E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})[/tex]

remember that [tex]q=49Q/64[/tex]

then:

[tex]Q(2R-x^{2})=49/64*x^{2}[/tex]

solving:

[tex]x_{1} =16R/15[/tex]

[tex]x_{2} =16R[/tex]

but: R<x<2R

so : [tex]x =16R/15 [/tex]