Home
Class 12
PHYSICS
A charge q is placed at the some distanc...

A charge `q` is placed at the some distance along the axis of a uniformly charged disc of surface charge density `sigma`. The flux due to the charge `q` through the disc is `phi`. The electric force on charge `q` exerted by the disc is

A

`sigma phi`

B

`(sigma phi)/(4pi)`

C

`(sigma phi)/(2pi)`

D

`(sigma phi)/(3 pi)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the electric force on charge \( q \) exerted by the uniformly charged disc with surface charge density \( \sigma \), we can follow these steps: ### Step 1: Understand the Problem We have a charge \( q \) placed at a distance from a uniformly charged disc with surface charge density \( \sigma \). The electric flux \( \phi \) through the disc due to the charge \( q \) is given. ### Step 2: Relate Electric Flux to Charge The electric flux \( \phi \) through the disc due to the charge \( q \) can be expressed using Gauss's Law: \[ \phi = \frac{q}{2 \epsilon_0} (1 - \cos \theta) \] where \( \theta \) is the half-angle subtended by the disc at the charge \( q \). ### Step 3: Find the Electric Field Due to the Disc The electric field \( E \) at a distance \( x \) from the center of the disc can be derived from the surface charge density \( \sigma \): \[ E = \frac{\sigma}{2 \epsilon_0} (1 - \frac{x}{\sqrt{x^2 + R^2}}) \] where \( R \) is the radius of the disc and \( x \) is the distance from the disc to the charge \( q \). ### Step 4: Calculate the Electric Force on Charge \( q \) The electric force \( F \) on the charge \( q \) due to the electric field \( E \) from the disc is given by: \[ F = qE \] Substituting the expression for \( E \): \[ F = q \cdot \frac{\sigma}{2 \epsilon_0} \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right) \] ### Step 5: Simplify the Expression Now, we can simplify the expression for the electric force: \[ F = \frac{q \sigma}{2 \epsilon_0} \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right) \] ### Final Result Thus, the electric force on charge \( q \) exerted by the disc is: \[ F = \frac{q \sigma}{2 \epsilon_0} \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right) \]

To find the electric force on charge \( q \) exerted by the uniformly charged disc with surface charge density \( \sigma \), we can follow these steps: ### Step 1: Understand the Problem We have a charge \( q \) placed at a distance from a uniformly charged disc with surface charge density \( \sigma \). The electric flux \( \phi \) through the disc due to the charge \( q \) is given. ### Step 2: Relate Electric Flux to Charge The electric flux \( \phi \) through the disc due to the charge \( q \) can be expressed using Gauss's Law: \[ ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ELECTRIC CHARGE, FIELD & FLUX

    A2Z|Exercise Section B - Assertion Reasoning|25 Videos
  • ELECTRIC CHARGE, FIELD & FLUX

    A2Z|Exercise AIPMTNEET Questions|24 Videos
  • ELECTRIC CHARGE, FIELD & FLUX

    A2Z|Exercise Electric Dipole|29 Videos
  • DUAL NATURE OF RADIATION AND MATTER

    A2Z|Exercise Section D - Chapter End Test|30 Videos
  • ELECTRIC POTENTIAL & CAPACITANCE

    A2Z|Exercise Section D - Chapter End Test|29 Videos

Similar Questions

Explore conceptually related problems

Field Of Uniformaly Charged Disc

Find the electric field on the axis of a charged disc.

Knowledge Check

  • A point charge placed on the axis of a uniformly charged disc experiences a force f due to the disc. If the charge density on the disc sigam is, the electric flux through the disc, due to the point charge will be

    A
    `(2 pi f)/(sigma)`
    B
    `(f)/(2 pi sigma)`
    C
    `(f^(2))/(sigma)`
    D
    `(f)/(sigma)`
  • A charge 'Q' is placed at the centre of a hemispherical surface of radius 'R'. The flux of electric field due to charge 'Q' through the surface of hemisphere is

    A
    `Q//4epsilon_(0)`
    B
    `Q//4piepsilon_(0)`
    C
    `Q//2epsilon_(0)`
    D
    `Q//2piepsilon_(0)`
  • Similar Questions

    Explore conceptually related problems

    A point charge q is placed at a distance d form the center of circular disc of radius R . Find electric flux through the disc due to that charge

    A uniformly charged disc of radius R having surface charge density sigma is placed in the xy plane with its center at the origin. Find the electric field intensity along the z-axis at a distance Z from origin :

    A charge q is placed at the centre of an imaginary spherical surface. What will be the electric flux due to this charge through any half of the sphere.

    Find the electric potential at the axis of a uniformly charged disc and use potential to find the electric field at same point.

    Calculate electric field at a point on axis, which at a distance x from centre of uniformly charged disc having surface charge density sigma and R which also contains a concentric hole of radius r.

    The diagram shows a part of disc of radius R carrying uniformly distributed charge of density sigma . Electric potential at the centre O of parent disc is