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A wire having a uniform linear charge de...

A wire having a uniform linear charge density `lambda` is bent in the form of a ring of radius `R`. Point `A` as shown in (Fig. 3.145) is in the plane of the ring but not at the center. Two elements of the ring of lengths `a_1` and `a_2` subtend very small same angle at point `A`. they are at distances `r_1` and `r_2` from point `A`, respectively `(r_2 gt r_1)`.
.

A

The ratio of charges of element `a_1` to that of element `a_2` is `r_1//r_s`.

B

The element `a_1` produced greater magnitude of electric field at `A` than elements `a_2`.

C

The same elements `a_1 and a_2` produce same potential at `A`

D

The direction of the net electric field produced by the elements only at `A` is toward element `a_2`.

Text Solution

Verified by Experts

The correct Answer is:
a.,b.,c.,d.

Charge on `a_1` is `(r_1 theta) lambda`
Charge on `a_2` is `(r_2 theta) lambda`
ratio of charged is `r_1//r_2`
`E_1 ("field produced by" a_1) = (K [(r_1 theta) lambda])/(r_1^2) = (K theta lambda)/r_(1)`
`E_2 ( "field produced by" a_2) = (K theta lambda)/r_(1)`
As`r_2 gt r_1, E_1 gt E_2` i.e., net field at `A` is toward `a_2`
`V_1 (K(r_1 theta))/r_(1) = K theta lambda`
`V_2 = (K(r_2 theta))/r_(1) = K theta lambda`
`V_1 = V_2`.
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Knowledge Check

  • A half ring of radius r has a linear charge density lambda .The potential at the centre of the half ring is

    A
    `(lambda)/(4epsioln_(0))`
    B
    `(lambda)/(4pi^(2)epsilon_(0)r)`
    C
    `(lambda)/(4piepsilon_(0)r)`
    D
    `(lambda)/(4piepsilon_(0)r^(2))`
  • A rod with linear charge density lambda is bent in the shap of circular ring. The electric potential at the centre of the circular ring is

    A
    `(lambda)/(epsilon_(0))`
    B
    `(lambda)/(2epsilon_(0))`
    C
    `(lambda)/(epsilon_(0))`
    D
    `(2lambda)/(epsilon_(0))`
  • An infinitely long wire with linear charge density +lambda is bent and the two parts are inclined at angle 30^@ as shown in fig, the electric field at point P is :

    A
    `(lambda)/(4 pi in_0 d) [ (1 + sqrt(3))hati + hatj]`
    B
    `(lambda)/(4 pi in_0 d) [ (1 - sqrt(3))hati + 2hatj]`
    C
    `(lambda)/(4 pi in_0 d) [ (1 + sqrt(3))hati - 2hatj]`
    D
    `(lambda)/(4 pi in_0 d) [ (1 + sqrt(3))hati + 2hatj]`