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Step-by-Step Solution
Step 1: Divide the Disc into Elemental Rings
Consider the uniformly charged disc lying in the xy-plane with radius $R$ and surface charge density $\sigma$. To find the electric field along the z-axis at distance $Z$ from the origin, divide the whole disc into thin concentric rings of radius $r$ and infinitesimal thickness $dr$.
Step 2: Compute the Area and Charge of the Elemental Ring
The area of a thin ring at radius $r$ with thickness $dr$ is given by:
$ dA = 2\pi r\,dr $
Hence, the charge on this elemental ring is:
$ dq = \sigma \, dA = \sigma \, (2\pi r\,dr). $
Step 3: Calculate the Contribution to the Electric Field from the Elemental Ring
We focus on the component of the electric field along the z-axis (say $dE_z$) due to this elemental ring. By symmetry, only the z-components of the field from each ring add up; the horizontal components cancel out. Using Coulomb’s law in differential form:
$ dE_z = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{dq \, Z}{\left(r^2 + Z^2\right)^{3/2}}.
Step 4: Substitute the Expression for dq
Replace $dq$ in the expression:
$ dE_z
= \frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma (2\pi r \, dr) \, Z}{(r^2 + Z^2)^{3/2}}
= \frac{\sigma Z}{2 \varepsilon_0} \cdot \frac{r\,dr}{(r^2 + Z^2)^{3/2}}.
Step 5: Integrate from r = 0 to r = R
To find the total electric field $E_z$ at the point on the z-axis, integrate the expression for $dE_z$ over $r$ from $0$ to $R$:
$ E_z
= \int_{0}^{R} \frac{\sigma Z}{2 \varepsilon_0} \cdot \frac{r\,dr}{(r^2 + Z^2)^{3/2}}.
The integral can be solved using a standard result or substitution method (e.g., setting $r^2 + Z^2 = u$). After evaluating:
$ E_z
= \frac{\sigma}{2\varepsilon_0} \left( 1 - \frac{Z}{\sqrt{Z^2 + R^2}} \right).
Step 6: Final Expression for the Electric Field
Thus, the magnitude of the electric field along the z-axis at a distance $Z$ from the center of the uniformly charged disc is:
$ E = \frac{\sigma}{2\varepsilon_0} \left( 1 - \frac{Z}{\sqrt{Z^2 + R^2}} \right).
Reference Diagram
Key Observations
The electric field decreases with increasing $Z$. Far from the disc ($Z \gg R$), the field resembles that of a point charge.
At $Z=0$, the field is $ \frac{\sigma}{2 \varepsilon_0} $, which is a well-known result for an infinite plane, modified slightly for a finite disc.
