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Question

Calculate electric field intensity due to uniformly charged non-conducting sphere:

(a) outside the sphere

(b) at the surface of sphere

(c)inside the sphere

(d) at the centre of the sphere

Plot at graph between electric Held intensity and distance.

(a) outside the sphere

(b) at the surface of sphere

(c)inside the sphere

(d) at the centre of the sphere

Plot at graph between electric Held intensity and distance.

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Solution

Electric field intensity due to a uniformly charged non-conducting sphere

when charge is given to non-conducting sphere , it uniformly spreads through out its volume . if q is the charge given and R is the radius of the sphere, then the volume charge density.

p=q43πR3

or p=3q4πR3............(1)

(a) Outside the sphere : in this case taking O as center and r as radius a spherical Gaussian surface is drawn . the point P will be situated at this surface.

the direction of →E will be outwards directed due to symmetry . the charge enclosed by the Gaussian surface.

Σq=q

Therefore from Gauss's law

ϕE=qϵ0..........(2)

From the defination of the electric flux,

ϕE=∮S→E→ds=∮SEdscos00=∮SEds

or ϕE=E∮Sds ( as E will be same at each points of Gaussian surface )

or ϕE=E4πr2...........(3)

Now equating equations (2) and (3) we get

E×4πr2=qϵ0

or E=14πϵ0qr2..........(4)

From equation (1)

p=3q4πR2⇒q=43πR3p

∴ From equation (4)

E=14πϵ0 frac4πR3p3r2

or E=pR33ϵ0r2

or E=p3ϵ0(R3r2)...........(5)

in vector form

→E=P3ϵ0(R3r2)^r

→E=p3ϵ0(R3r2)^r...........(6)

where ^r = unit vector in direction OP

(b) When point is situated on the surface of the sphere ; in this case we can consider the surface of the surface as Gaussian surface . therefore the whole charge of the sphere will again be charge enclosed.thus by putting r=R equation we get

E=P3ϵ0(R3R2)

or E=pR3ϵ0..........(7)

In vector from

→E=pR3ϵ0^r...........(8)

(c) When point P is inside the sphere (r<R)

Here also we draw a spherical Gaussian surface taking O as center and r as radius . point P will be situated at this surface. now the charge enclosed by this Gaussian surface.

Σq=p43πr3

Substituting the value of p form equation (1)

σq=3q4πR3×4πr33

or Σq=qr3R3

Therefore from Gauss's law,

ϕE=Σqϵ0=qr3ϵ0R3

or ϕE=qr3ϵ0R3..........(9)

From the defination of the electric flux,

ϕE=∮S→E→ds=∮SEdscos00

(for spherical surface )

=E∮Sds

or ϕE=E×πr2........(10)

Now equating equation (9) and (10) we have

E×4πr2=qr3ϵ0R3

∴E=14πϵ0qr3r2R3

or E=14πϵ0qrR3.........(11)

Again from the equation (1)

q=43πR3p

∴ From equation (11)

E=14πϵ0rR3×43πR3p

or E=p3ϵ0r .............(12)

In vector form ,

→E=p3ϵ0r^r...........(13)

(d) At the centre of the sphere : for central point

r=0

∴ From equation (12)

E=0.........(14)

The sketch showing the variation of E with r is given figure

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