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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: \[ ...
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