Question

The electrostatic potential inside a charged sphere is given as V = Ar² + B, where r is the distance from the centre of the sphere; A and B are constants. Then the charge density in the sphere is ______.

विकल्प

• 16 A`epsilon_0`

• -6 A`epsilon_0`

• 20 A`epsilon_0`

• -15 A`epsilon_0`

Answer 1

The electrostatic potential inside a charged sphere is given as V = Ar² + B, where r is the distance from the centre of the sphere; A and B are constants. Then the charge density in the sphere is -6 A`epsilon_0`.

Explanation:

E= `-"dV"/"dr" = -"d"/"dr"("A""r"^2 + "B") = -2"A""r"`  ... (i)

But E= `1/(4piepsilon_0) = "q"/"r"^2 = ((4/3pi"r"^3)"p")/(4piepsilon_0"r"^2) ="pr"/(3epsilon_0` .... (ii)

From equations (i) and (ii),

`"pr"/(3epsilon_0) = -2"Ar"`

`therefore "p" = -6"A"epsilon_0`