Question

Two 2.4 cm -diameter disks face each other, 1.0 mm apart. They are charged to ±11nC. What is the electric field strength between the disks?

Answer 1

Answer:

$\rm 1.374\times 10^6\ N/C.$

Explanation:

Given:

• Diameter of each disc, D = 2.4 cm.
• Distance between the discs, d = 1.0 mm.
• Charges on the discs, q = ±11 nC = $\rm \pm 11\times 10^{-9}\ C.$

The surface area of each of the disc is given by

$\rm A=\pi \times Radius^2.\\\\Radius = \dfrac D2=\dfrac{2.4\ cm}{2} = 1.2\ cm = 1.2\times 10^{-2}\ m.\\A = \pi \times (1.2\times 10^{-2} )^2=4.524\times 10^{-4}\ m^2.$

For the case, when d<<D, the strength of the electric field at a point due to a charged sheet is given by

$\rm E=\dfrac{|\sigma|}{2\epsilon_o}$

where,

• $\sigma$ = surface charge density of the disc = $\rm \dfrac qA$.

• $\epsilon_o$ = electrical permittivity of free space = [tex\rm ]9\tiimes 10^9\ Nm^2/C^2[/tex].

The electric field strength between the discs due to negative disc is given by

$\rm E_1 = \dfrac{|\sigma|}{2\epsilon_o}=\dfrac{|q|}{2A\epsilon_o}.$

Since, the electric field is directed from positive charge to negative charge, therefore, the direction of this electric field is towards the negative disc.

The electric field strength between the discs due to positive disc is given by

$\rm E_2 = \dfrac{|\sigma|}{2\epsilon_o}=\dfrac{|q|}{2A\epsilon_o}.$

The direction of this electric field is away from the positive disc, i.e., towards negative disc.

The electric field between the discs due to both the disc is towards the negative disc, therefore the total electric field strength between the discs is given by

$\rm E=E_1+E_2\\=\dfrac{|q|}{2A\epsilon_o}+\dfrac{|q|}{2A\epsilon_o}\\=\dfrac{|q|}{A\epsilon_o}\\=\dfrac{11\times 10^{-9}}{2\times (4.524\times 10^{-4})\times (8.85\times 10^{-12})}\\=1.374\times 10^6\ N/C.$