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If the total charge on a conducting hemisphere of radius R is Q, then the surface charge density will be:
1. \(\frac{Q}{3\pi R^2}\)
2. \(\frac{Q}{4\pi R^2}\)
3. \(\frac{Q}{2\pi R^2}\)
4. \(\frac{Q}{\pi R^2}\)

1 Answer

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Correct Answer - Option 1 : \(\frac{Q}{3\pi R^2}\)

CONCEPT:

Linear charge density:

  • It is defined as the quantity of charge per unit length.
  • Its SI unit is C/m.
  • If ΔQ charge is contained in the line element Δl, the linear charge density λ will be,

\(⇒ \lambda=\frac{Δ Q}{Δ l}\)

Surface charge density:

  • It is defined as the quantity of charge per unit area.
  • Its SI unit is C/m2.
  • If ΔQ charge is contained in the elemental area Δs, the surface charge density σ will be,

\(⇒ \sigma=\frac{Δ Q}{Δ s}\)

Volume charge density:

  • It is defined as the quantity of charge per unit volume.
  • Its SI unit is C/m3.
  • If ΔQ charge is contained in the elemental volume Δv, the volume charge density ρ will be,

\(⇒ \rho=\frac{Δ Q}{Δ v}\)

EXPLANATION:

Given ΔQ = Q

  • We know that in a conducting body the charge always resides at the surface because the charge can flow in a conducting body.
  • So all the charge Q will remain at the surface of the hemisphere.

We know that the total surface area of the hemisphere of radius R is given as,

⇒ Δs = 3πR2       ----(1)

If ΔQ charge is contained in the elemental area Δs, the surface charge density σ will be,

\(⇒ \sigma=\frac{Δ Q}{Δ s}\)       ----(2)

By equation 1 and equation 2 the surface charge density is given as,

\(⇒ \sigma=\frac{Q}{3πR^2}\)

  • Hence, option 1 is correct.

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