Question

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. \(E = \frac{\sigma}{2\epsilon_0} \left( 1 + \frac{Z}{(Z^2 + R^2)^{1/2}} \right)\)
• B. \(E = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{Z}{(Z^2 + R^2)^{1/2}} \right)\)
• C. \(E = \frac{2\epsilon_0}{\sigma} \left( \frac{1}{(Z^2 + R^2)^{1/2}} + Z \right)\)
• D. \(E = \frac{\sigma}{2\epsilon_0} \left( \frac{1}{(Z^2 + R^2)} + \frac{1}{Z^2} \right)\)

Hint:

It is highly recommended to memorize the electric field formulas for standard charge distributions like a ring, disc, infinite line, and infinite sheet. For the disc, you can check the formula by considering two limits: 1. When \(Z \to 0\) (close to the center), \(E \to \frac{\sigma}{2\epsilon_0}\) (field of an infinite sheet). 2. When \(Z \gg R\) (far away), the disc looks like a point charge. The formula approximates to \(E \approx \frac{1}{4\pi\epsilon_0} \frac{Q}{Z^2}\), where \(Q = \sigma \pi R^2\). The given formula satisfies both these limits.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the formula for the electric field at a point on the axis of a uniformly charged circular disc. This is a standard result in electrostatics, derived using integration.
Step 2: Key Formula or Approach:
The electric field of a charged disc is found by integrating the contributions from infinitesimal charged rings that make up the disc. The electric field dE due to a ring of radius \(r\) and charge \(dq\) at a point \(Z\) on its axis is \(dE = \frac{1}{4\pi\epsilon_0} \frac{Z dq}{(r^2 + Z^2)^{3/2}}\). We express \(dq\) in terms of the surface charge density \(\sigma\) (\(dq = \sigma dA = \sigma(2\pi r dr)\)) and integrate from \(r=0\) to \(r=R\).
\[ E = \int_0^R \frac{1}{4\pi\epsilon_0} \frac{Z (\sigma 2\pi r dr)}{(r^2 + Z^2)^{3/2}} = \frac{\sigma Z}{2\epsilon_0} \int_0^R \frac{r dr}{(r^2 + Z^2)^{3/2}} \] Step 3: Detailed Explanation (Derivation):
Let's perform the integration. Let \(u = r^2 + Z^2\), so \(du = 2r dr\), or \(r dr = du/2\).
The limits of integration change from \(r=0 \to u=Z^2\) and \(r=R \to u=R^2+Z^2\). \[ E = \frac{\sigma Z}{2\epsilon_0} \int_{Z^2}^{R^2+Z^2} \frac{du/2}{u^{3/2}} = \frac{\sigma Z}{4\epsilon_0} \int_{Z^2}^{R^2+Z^2} u^{-3/2} du \] \[ E = \frac{\sigma Z}{4\epsilon_0} \left[ \frac{u^{-1/2}}{-1/2} \right]_{Z^2}^{R^2+Z^2} = \frac{\sigma Z}{4\epsilon_0} \left[ -2u^{-1/2} \right]_{Z^2}^{R^2+Z^2} \] \[ E = -\frac{\sigma Z}{2\epsilon_0} \left[ \frac{1}{\sqrt{u}} \right]_{Z^2}^{R^2+Z^2} = -\frac{\sigma Z}{2\epsilon_0} \left( \frac{1}{\sqrt{R^2+Z^2}} - \frac{1}{\sqrt{Z^2}} \right) \] Assuming \(Z>0\), \(\sqrt{Z^2} = Z\). \[ E = -\frac{\sigma Z}{2\epsilon_0} \left( \frac{1}{\sqrt{R^2+Z^2}} - \frac{1}{Z} \right) = \frac{\sigma}{2\epsilon_0} \left( - \frac{Z}{\sqrt{R^2+Z^2}} + 1 \right) \] \[ E = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{Z}{\sqrt{Z^2 + R^2}} \right) \] Step 4: Final Answer:
The derived formula is \(E = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{Z}{(Z^2 + R^2)^{1/2}} \right)\). This is a standard result and matches option (B).