Home
Class 12
PHYSICS
A ball of radius R carries a positive ch...

A ball of radius R carries a positive charges whose volume density at a point is given as `rho=rho_(0)(1-r//R)`, Where `rho_(0)` is a constant and r is the distance of the point from the center. Assume the permittivities of the ball and the enviroment to be equal to unity.
(a) Find the magnitude of the electric field strength as a function of the distance r both inside and outside the ball.
(b) Find the maximum intensity `E_("max")` and the corresponding distance `r_(m)`.

Text Solution

Verified by Experts

(a) We assume the ball to be divided into infinite number of concentric thin shells of thickness dr. Let us assume such a shell at a radial distance `'r'`.
Volume of this shell = `dV=4pir^(2)dr`
Charge in the shell = `rhodV=4pir^(2)rhodr=4pir^(2)rho_(0)(1-(r)/(R))dr`
Net charge enclosed in the sphere of radius 'r'
`q=intrhodv=underset(0)overset(r)int4pirho_(0)r^(2)(1-(r)/(R))dr=4pirho_(o)((r^(3))/(3)-(r^(4))/(4R))`
Electric field at radial distance of 'r'
`E=(q)/(4piepsilon_(0)r^(2))=(1)/(4piepsilon_(0)r^(2))xx4pirho_(0)((r^(3))/(3)-(r^(4))/(4R))=(rho_(0)r)/(3epsilon_(0))(1-(3r)/(4R))`
Now for those points outside the sphere, we have to take into account the total charges contained in the sphere of radius R.
Total charge `Q=underset(0)overset(R)intrho_(0)(1-(r)/(R))4pir^(2)dr=(pirho_(0)R^(3))/(3)`
Electric field, using Gauss's law, at a point which is at a distance `r(gtR)` from the centre,
`intE.ds=(Q)/(epsilon_(0))impliesE=(Q)/(4piepsilon_(0)r^(2))=(rho_(0)R^(3))/(12epsilon_(0)r^(2))`

(b) Again let us consider the expression for electric field within the sphere of radius R:
`E=(rho_(0)r)/(3epsilon_(0))(1-(3r)/(4R))`.
For maximum electric field:
`(dE)/(dr)=0implies1-(6r)/(4R)=0impliesr=(2R)/(3)`
Putting, `r=(2R)/(3)`, value of maximum electric field, `E_(max)=(rho_(0)R)/(9in_(0))`,
Promotional Banner

Similar Questions

Explore conceptually related problems

A thin wire of radius ..r.. carries a charge q. Find the magnitude of the electric field strength on the axis of the ring as a function of distance L from the centre. Find the same for L gt gt r Find maximum field strength and the corresponding distance L.

A sphere of charges of radius R carries a positive charge whose volume chasrge density depends only on the distance r from the ball's centre as rho=rho_0(1-r/R), where rho_0 is constant. Assume epsilon as theh permittivity of space. the maximum electric field intensity is

A ball of radius R carries a positive charge throughout its volume, such that the volume density of charge depends on distance r from the ball's centre as rho = rho _(0) ( 1- ( r )/®) , where rho_(0) is a constant. Assuming the permittivity of ball to be one , find magnitude of electric field as a function of distance r, both inside and outside the ball. Strategy : The field has a spherical symmetry . For a point outside the ball ( r gt R ) phi =oint vec(E). bar(dA) = E xx 4 pi r^(2) By Gauss law , phi= ( Q)/( epsilon_(0)) implies E_(out) = (Q)/( 4pi epsilon_(0)r^(2)) , where Q is total charge For a point inside the ball ( r lt R ) phi = oint vec(E) . bar(dA) = E xx 4 pi r^(2) By Gauss law, phi = ( q_(enc))/(epsilon_(0)) implies E_("in")= ( q_(enc))/( 4pi epsilon_(0)r^(2)) where q_(enc)= charge in a sphere of radius r ( lt R) To find the enclosed charge , consider a spherical shell of radius r and thickness dr Its volume dV = 4pi r^(2) dr Charge dq = rho dV implies dq= rho _(0) (1-(r )/(R)) 4pi r^(2) dr implies q= int _(0)^(r ) rho _(0) ( 1- (r)/( R )) 4pi r^(2) dr = rho _(0)4pi [ int_(0)^(r ) r^(2) dr - int_(0)^(r ) (r^(3))/(R) dr ]= rho _(0) 4pi [ ( r^(3))/( 3) - (r^(4))/( 4R)]

A sphere of charges of radius R carries a positive charge whose volume chasrge density depends only on the distance r from the ball's centre as rho=rho_0(1-r/R), where rho_0 is constant. Assume epsilon as theh permittivity of space. The magnitude of electric field as a function of the distance r inside the sphere is given by

A sphere of charges of radius R carries a positive charge whose volume charge density depends only on the distance r from the ball's centre as rho=rho_0(1-r/R) , where rho_0 is constant. Assume epsilon as theh permittivity of space. The magnitude of the electric field as a functiion of the distance r outside the balll is given by

A sphere of charges of radius R carries a positive charge whose volume chasrge density depends only on the distance r from the ball's centre as rho=rho_0(1-r/R) , where rho_0 is constant. Assume epsilon as theh permittivity of space. The value of distance r_m at which electric field intensity is maximum is given by

A non-conducting solid sphere has volume charge density that varies as rho=rho_(0) r, where rho_(0) is a constant and r is distance from centre. Find out electric field intensities at following positions. (i) r lt R" " (ii) r ge R

A solid ball of radius R has a charge density rho given by rho = rho_(0)(1-(r )/(R )) for 0le r le R The electric field outside the ball is :

A solid non conducting sphere of radius R has a non-uniform charge distribution of volume charge density, rho=rho_(0)r/R , where rho_(0) is a constant and r is the distance from the centre of the sphere. Show that : (i) the total charge on the sphere is Q=pirho_(0)R^(3) (ii) the electric field inside the sphere has a magnitude given by, E=(KQr^(2))/R^(4)

The density inside a solid sphere of radius a is given by rho=rho_0/r , where rho_0 is the density ast the surface and r denotes the distance from the centre. Find the graittional field due to this sphere at a distance 2a from its centre.