Home
Class 12
PHYSICS
The maximum electric field upon the axis...

The maximum electric field upon the axis of a circular ring `( q,R)` is given by `E = ( q)/( pi epsilon_(0)R^(2))xx(1)/( 6 sqrt(n))` . Find n.

Text Solution

Verified by Experts

The correct Answer is:
3
Promotional Banner

Topper's Solved these Questions

  • ELECTRIC CHARGES AND FIELDS

    AAKASH INSTITUTE|Exercise SECTION-H( MULTIPLE TRUE-FALSE QUESTIONS)|10 Videos
  • ELECTRIC CHARGES AND FIELDS

    AAKASH INSTITUTE|Exercise SECTION-I(SUBJECTIVE TYPE QUESTIONS)|2 Videos
  • ELECTRIC CHARGES AND FIELDS

    AAKASH INSTITUTE|Exercise SECTION-F ( Matrix-Match Type Questions)|5 Videos
  • DUAL NATURE OF RADIATION AND MATTER

    AAKASH INSTITUTE|Exercise ASSIGNMENT (SECTION-D)|10 Videos
  • ELECTROMAGNETIC INDUCTION

    AAKASH INSTITUTE|Exercise ASSIGNMENT(SECTION -D) Assertion-Reason type Question)|15 Videos

Similar Questions

Explore conceptually related problems

The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R will be

The electric field strength due to a ring of radius R at a distance x from its centre on the axis of ring carrying charge Q is given by E = (1)/(4 pi epsilon_(0)) (Qx)/((R^(2) + x^(2))^(3//2)) At what distance from the centre will the electric field be maximum ?

The electric field at point P due to a charged ball is given by E_(p)=(1)/( 4pi epsilon_(0))(q)/(r^(2)) To measure 'E' at point P, A test charge q_(0) is placed at point P and measure electric force F upon the test charge. Check whether (F)/(q_(0)) is equal to (1)/(4pi epsilon_(0))(q)/(r^(2)) or not .

The maximum electric field at a point on the axis of a uniformly charged ring is E_(0) . At how many points on the axis will the magnitude of the electric field be E_(0)//2 .

The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R is at a distance x from the centre of the ring.The value of x is

For spherical symmetrical charge distribution, variation of electric potential with distance from centre is given in diagram Given that : V=(q)/(4pi epsilon_(0)R_(0)) for r le R_(0) and V=(q)/(4 piepsilon_(0)r) for r ge R_(0) Then which option (s) are correct : (1) Total charge within 2R_(0) is q (2) Total electrosstatic energy for r le R_(0) is non-zero (3) At r = R_(0) electric field is discontinuous (4) There will be no charge anywhere except at r lt R_(0)

If the elctron ( charge of each electron =-e) are shifted by a small distance x,a net + ve charge density ( per unit area ) is induced on the surface. This will result in an electric field E=n ex //epsilon_(0) in the direction of x and a restoring force on an electron of -(n e^(2)x)/(epsilon_(0)) , Thus m ddot(x) =-(n e^(2)x)/(epsilon_(0)) or ddot(x) + ( n e^(2))/( m epsilon_(0))x=0 This gives omega_(p)= sqrt((n e^(2))/( m epsilon_(0)))=1.645xx10^(16)s^(-1) as the plasma frequency for the problem,.

AAKASH INSTITUTE-ELECTRIC CHARGES AND FIELDS -SECTION-G(INTEGER ANSWER TYPE QUESTIONS)
  1. The maximum electric field upon the axis of a circular ring ( q,R) is ...

    Text Solution

    |