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An infinitely long solid cylinder of radius R carries a uniform volume charge density rho. A spherical cavity of radius R/2 is carved out with its centre on the cylinder axis. The magnitude of the electric field at point P, located a distance 2R from the axis, is found to equal 23*rho*R/(16*k*epsilon0). Find the value of k.

1. 6
2. 1
3. 2
4. 4

Correct answer: 6

Solution

By superposition, E_P = E_full_cylinder - E_removed_sphere (the cavity = adding negative charge). For the full cylinder at d = 2R: E_cyl = rho*R²/(2*epsilon0*(2R)) = rho*R/(4*epsilon0). The removed sphere of radius R/2 has charge magnitude rho*(4/3)pi(R/2)³ = rho*pi*R³/6; treated as a point at distance 2R, E_sph = (1/(4*pi*epsilon0)) * (rho*pi*R³/6)/(2R)² = rho*R/(96*epsilon0). The two fields point oppositely (cavity removes positive charge), so net E = rho*R/(4*epsilon0) - rho*R/(96*epsilon0) = (24 - 1)*rho*R/(96*epsilon0) = 23*rho*R/(96*epsilon0). Writing 96 = 16*6 gives 23*rho*R/(16*6*epsilon0), so k = 6.

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