Question

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

Answer 1

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See  attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$  (1)

according to the law of gauss and superposition Law:

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

$E_{Q}=kq_{2}/(r_{1}^{2})=$

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

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

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

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

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

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

$q=49Q/64$

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

if R<x<2R   AND $E_{total}=E_{Q}-E_{q}=0$

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

remember that  $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R  

so : $x =16R/15$