IIT JEE , AIEEE Forum - Physics , Electricity

ackhill

In region der exists an electric field E(bar)=(4x^3i+3y^2j+8zk)\(4x^4+3y^3+8z^2).
the charge enclosed within a sphere of radius R cntered at origin(x,y,z=0) is
a)2pie(epsilon not) R
b)4pie(epsilon not) R
c)(4x^3+3y^2+8z)(epsilon not)
d)zero..

narayana

i think its d man

electric field inside a spherical conductor is zero

gcch29

at any point P(x,y,z) on sphere a unit vector prependicular to sphere radially outwards is = (xi +yj+zk)/(x²+ y² +z² )^1/2
x^2+y^2+z^2 =r^2
d

= E.(unit vector)ds
integrate both sides
on intgerating and evaluating dot product

=4

r^2/r
q/

=4

r
q=4

r

ackhill

@narayan...no yaar its true dat electric field inside a conductor is zero..but here charge enclosed is asked n hence we`ve 2 use gauss law n first find flux through d sphere..

magiclko

the flux enclosed is given by d

= E.ds (E and s both are vector quantities)

ds = n ds , whr n is the unit normal vector, and is given by

n =

(S) / |

(S)|, whr

is the del operator, and is given by

= d/dx i + d/dy j + d/dz k

i.e.

(S) = dS/dx i + dS/dy j + dS/dz k

and S = x²+ y² +z² -r²

thrfore, n = (xi +yj+zk) / (x²+ y² +z² )^1/2

= (xi +yj+zk) / r

thus, E.ds = E.n ds

and, E.n =((4x³i + 3y²j + 8zk) / (4x⁴ +3y³+ 8z²)) . (xi +yj+zk) / r

= (4x⁴ +3y³+ 8z²)

r (4x⁴ +3y³+ 8z²)

= 1/r

thrfore, d

= E.ds

= 1/r

ds

q/

=4

r
q=4

r

PS: @gcch29, the answer was undoubtedly correct, but this is the correct procedure, not the one which u have given...

moreover, this method wont come in JEE, cuz it involves the funda of 'surface integral' which is defintely nt in JEE syllabus...

Ask & Discuss Questions with Community & Experts