An infinitely long solid cylinder of radius has a uniform volume charge density . It has a spherical cavity of radius with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point , which is at a d

Question

An infinitely long solid cylinder of radius $R$ has a uniform volume charge density $ρ$ . It has a spherical cavity of radius $R / 2$ with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point $P$ , which is at a distance $2 R$ from the axis of the cylinder, is given by the expression $\frac{23 ρ R}{16 k ε_{0}} .$ The value of $k$ is [2012]

Smart Achievers · Accepted Answer

(6)

The magnitude of the electric field at the point P which is at a distance 2R from the axis of the cylinder

$E = E_{total} - E_{cavity}$

$= \frac{γ}{2 π ε_{0} (2 R)} - \frac{1}{4 π ε_{0}} \frac{Q}{{(2 R)}^{2}}$

$Q_{sphere} = \frac{4}{3} π {(\frac{R}{2})}^{3} ρ = \frac{π R^{3} ρ}{6};$ $λ_{cylinder} = π R^{2} ρ$

$∴$ $E = \frac{π R^{2} ρ}{4 π ε_{0} R} - \frac{1}{4 π ε_{0}} \cdot \frac{π R^{3} ρ / 6}{4 R^{2}}$

$= \frac{23 ρ R}{96 ε_{0}} = \frac{23 ρ R}{16 \times 6 \times ε_{0}}$

$∴$ $k = 6$

Solution

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Solution

Ans. (6) The magnitude of the electric field at the point P which is at a distance 2R from the axis of the cylinder E=Etotal-Ecavity =γ2πε0(2R)-14πε0Q(2R)2 Qsphere=43π(R2)3ρ=πR3ρ6; λcylinder=πR2ρ ∴ E=πR2ρ4πε0R-14πε0·πR3ρ/64R2 =23ρR96ε0=23ρR16×6×ε0 ∴ k=6

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