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The nuclear charge (Ze) is non uniformll...

The nuclear charge (`Ze`) is non uniformlly distribute with in a nucleus of radius r. The charge density `rho(r)` (charge per unit volume) is dependent only on the radial distance r form the centre of the nucleus s shown in figure. The electric field is only along the radial direction.

The electric field at `r=R` is

A

Independent of a

B

Directly proportional to a

C

Directly proportional to `a^(2)`

D

Inversely proportional to a

Text Solution

Verified by Experts

The correct Answer is:
A
`E(4pi R^(2))=((Ze))/in_(0)rArr E=(Ze)/(4pi in_(0) R^(2))`
`rArr` Independent of a
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Knowledge Check

  • The nuclear charge ( Ze ) is non uniformlly distribute with in a nucleus of radius r. The charge densilty rho(r) (charge per unit volume) is dependent only on the radial distance r form the centre of the nucleus s shown in figure. The electric field is only along the radial direction. The electric field within the nucleus is generaly observed to be linearly dependent on r. This implies

    A
    `a=0`
    B
    `a=R/2`
    C
    `a=R`
    D
    `a=(2R)/3`

  • The nuclear charge ( Ze ) is non uniformlly distribute with in a nucleus of radius r . The charge density rho(r) (charge per unit volume) is dependent only on the radial distance r form the centre of the nucleus s shown in figure. The electric field is only along the radial direction. For a=0 the value of d (maximum value of rho as shown in the figure) is

    A
    `(3Ze)/(4piR^3)`
    B
    `(3Ze)/(piR^3)`
    C
    `(4Ze)/(3piR^3)`
    D
    `(Ze)/(3piR^3)`

  • Charges q is uniformly distributed over a thin half ring of radius R . The electric field at the centre of the ring is

    A
    `(q)/(2pi^(2)epsilon_(0)R^(2))`
    B
    `(q)/(4pi^(2)epsilon_(0)R^(2))`
    C
    `(q)/(4piepsilon_(0)R^(2))`
    D
    `(q)/(2piepsilon_(0)R^(2))`

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