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

`q=49Q/64`

and

`x =16R/15`

Step-by-step 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`