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A rod lies along the X-axis with one end...

A rod lies along the X-axis with one end at the origin and the other at `x rarr oo`. It carries a uniform charge `lambda Cm^(-1)`. The electric field at the point `x=-a` on the axis will be

A

`E=lambda/(4pi epsi_(0)a) (-hat(i))`

B

`E=lambda/(4pi epsi_(0)a) (hat(i))`

C

`E=lambda/(2pi epsi_(0)a) (-hat(i))`

D

`E=lambda/(2pi epsi_(0)a) (hat(i))`

Text Solution

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The correct Answer is:
C
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Knowledge Check

  • A rod lies along the x-axis with one end at the origin and the other at x rarr oo . It carries a uniform charge lambda C/m. The electric field at the point x = -a on the axis will be

    A
    `vec(E) = (lambda)/(4pi epsilon_(0)a)(-hat(i))`
    B
    `vec(E) = (lambda)/(4pi epsilon_(0)a)(hat(i))`
    C
    `vec(E) = (lambda)/(2pi epsilon_(0)a)(-hat(i))`
    D
    `E = (lambda)/(2pi epsilon_(0)a)(hat(i))`

  • A positively charged rod lies along X-axis in such a manner that one end of the rod is at the origin and the other end at x =-oo . The linear charge density for the rod is lambda . Electric field intensity at x =a is

    A
    `lambda/(4piepsilon_0a)hati`
    B
    `-lambda/(4piepsilon_0a)hati`
    C
    `lambda/(4piepsilon_0a^2)hati`
    D
    `-lambda/(4piepsilon_0a^2) hati`

  • A rod of length l is placed along x - axis. One of its ends is at the origin. The rod has a non - uniform charge density lambda=(a)/(x) , a being a positive constant. The electric potential at the point P (origin) as shown in the figure is

    A
    `V=(a)/(4pi epsilon_(0))((l)/(b(b+l)))`
    B
    `V=(a)/(4pi epsilon_(0))((b)/(l(b+l)))`
    C
    `V=(a)/(4pi epsilon_(0))((b)/(l))`
    D
    `V=(a)/(4pi epsilon_(0))((l)/(b))`

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