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Q.

A cylindrical portion of radius r is removed from a solid sphere of radius R and uniform volume charge density $ρ$ in such a way that the axis of the hollow cylinder coincides with one of the diameters of the sphere. (r is negligible compared to R). Then the electric field intensity at point A is

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a

$- \frac{ρr}{6 ε_{0}} \hat{i}$

b

$\frac{ρr}{3 ε_{0}} \hat{i}$

c

$\frac{ρr}{6 ε_{0}} \hat{i}$

d

$- \frac{ρr}{3 ε_{0}} \hat{i}$

answer is C.

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Detailed Solution

Field at A due to the solid sphere without the cylindrical cavity
$E_{1} = - \frac{ρr}{3 ε_{0}} \hat{i}$
field at A due to the cylinder of length 2R (which can be assumed to be infinite, since r << R)
$E_{2} = \frac{2 K (ρ π r^{2})}{r} (- \hat{i}) = - \frac{ρ}{2 ε_{0}} r \hat{i}$
$∴ net field E = E_{1} - E_{2} = \frac{ρr}{6 ε_{0}} \hat{i}$

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A cylindrical portion of radius r is removed from a solid sphere of radius R and uniform volume charge density ρ in such a way that the axis of the hollow cylinder coincides with one of the diameters of the sphere. (r is negligible compared to R). Then the electric field intensity at point A is