91Ó°ÊÓ

• Radius of the disk,$R=2.5\text{ c³¾}$
• Surface density of the upper face of the disk,$\mathrm{σ}=5.3{\text{ μ°ä/³¾}}^{\text{2}}$
• Distance of the point from the disk,$z=12\text{ c³¾}$

Using the given formula of the electric field of a point due to a disk at a distance from it, we can get the required magnitude value of the field.

Formula:

The magnitude of the electric field produced by the disk at a point on its central axis, $E=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_o}(1−\frac{z}{\sqrt{z^2+R^2}})$ (i)

where,$\mathrm{σ}$= surface charge density

z = distanceon the central axis of the disk

R = Radius of the disk

The magnitude of the electric field produced by the disk at a point on its central axis is given using the equation (i) and the given data as follows:

$\begin{array}{c}E=\frac{(5.3×{10}^{−6}{\text{ C/³¾}}^{\text{2}})}{2(8.99×{10}^9\text{ N}â‹\dots \frac{{\text{m}}^2}{{\text{C}}^{\text{2}}})}[1−\frac{12\text{ c³¾}}{\sqrt{{\left(12\text{cm}\right)}^2+{\left(2.5\text{ c³¾}\right)}^2}}]\\ =6.3×{10}^3\text{ N/°ä}\end{array}$

Hence, the value of the electric field is $6.3×{10}^3\text{ N/°ä}$.