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Calculate the potential due to a thin un...

Calculate the potential due to a thin uniformly charged rod of length L at the poiot P shown in figure. The lioear charge density of the rod is `lambda C//m`

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Knowledge Check

  • The electrostatic potential due to the charge configuration at point P as shown in figure for bltlta is

    A
    `(2q)/(4piepsilon_0a)`
    B
    `(2qb^2)/(4piepsilon_0a^3)`
    C
    `(qb^2)/(4piepsilon_0a^3)`
    D
    zero

  • If linear charge density of a wire as shown in the figure is lambda

    A
    potential at the centre is `(lambda)/(2 epsi_(0))`
    B
    electric field at the centre of the loop is `(lambda)/(2 pi epsi_(0)R)`
    C
    electric field at the centre of the loop is `(lambda)/(2 pi epsi_(0)R) + (lambda)/(2 epsi_(0)R)`
    D
    None of the above

  • Find the electric field at point P (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length L carrying a charge Q. The distance of the point P from the centre of the rod is a =sqrt3/2L .

    A
    `(sqrt3Q)/(4piepsi_0L^2)`
    B
    `Q/(3piepsi_0L^2)`
    C
    `Q/(2sqrt3piepsi_0L^2)`
    D
    `Q/(4piepsi_0L^2)`

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    Find the electric field at a point P on the perpendicular bisector of a uniformly charged rod. The lengthof the rod is L, the charge on it is Q and the distance of P from the centre of the rod is a.

    Two uniform thin rods each of length L and mass m are joined as shown in the figure. Find the distance of centre of mass the system from point O

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    Charge Q is uniformly distributed on a dielectric rod AB of length 2l . The potential at P shown in the figure is equal to

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