A long conductor of circular cross section has inner radius r and outer radius 2r as shown in figure- Conductivity of material varies linearly with distance from axis- Conductivity of material is - on inner surface and 4- on outer surface- If electric field strength at a distance 3r2 from axis is E0- if constant potential difference V0 is applied across inner and outer surface- If electric field near the inner surface is nE0- Value of 4n- is

Σ( x )=σ +3σrx J = σ E J2π (r + x)l = constant 2πσ( 1+3xr)( r+x )E( x )=constant E_0( 1+3/2)( 32r )=E·( 1 )( r ) E=154E_0 =3.75 E_0 ...

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Standard VIII

Physics

Luminous Intensity

A long conduc...

Question

A long conductor of circular cross section has inner radius $r$ and outer radius $2 r$ as shown in figure. Conductivity of material varies linearly with distance from axis. Conductivity of material is $σ$ on inner surface and $4 σ$ on outer surface. If electric field strength at a distance $\frac{3 r}{2}$ from axis is $E_{0}$ , if constant potential difference $V_{0}$ is applied across inner and outer surface. If electric field near the inner surface is $n E_{0}$ . Value of $4 n$ , is

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Solution

$σ (x) = σ + \frac{3 σ}{r} x$
$J = σ E$
$J 2 π (r + x) l =$ constant
$2 π σ (1 + \frac{3 x}{r}) (r + x) E (x) = constant$
$E_{0} (1 + \frac{3}{2}) (\frac{3}{2} r) = E \cdot (1) (r)$
$E = \frac{15}{4} E_{0}$
$= 3.75 E_{0}$

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σ(x)=σ+3σrx J=σE J2π (r+x)l= constant 2πσ(1+3xr)(r+x)E(x)=constant E0(1+32)(32r)=E⋅(1)(r) E=154E0 =3.75 E0

$σ (x) = σ + \frac{3 σ}{r} x$
$J = σ E$
$J 2 π (r + x) l =$ constant
$2 π σ (1 + \frac{3 x}{r}) (r + x) E (x) = constant$
$E_{0} (1 + \frac{3}{2}) (\frac{3}{2} r) = E \cdot (1) (r)$
$E = \frac{15}{4} E_{0}$
$= 3.75 E_{0}$