Answer:
[tex]E=1.27\times 10^{12}\ N/C[/tex]
Explanation:
Given that,
Radius of the disk, r = 4.9 cm = 0.049 m
Charge, q = +5.6 C
We need to find the the electric field at a point on the axis and 3 mm from the center, x = 0.003 m
At a point on the axis of a ring, the electric field is given by :
[tex]E=\dfrac{kqx}{(x^2+r^2)^{3/2}}[/tex]
[tex]E=\dfrac{9\times 10^9\times 5.6\times 0.003}{(0.003^2+0.049^2)^{3/2}}[/tex]
[tex]E=1.27\times 10^{12}\ N/C[/tex]
So, the electric field at a point on the axis is [tex]1.27\times 10^{12}\ N/C[/tex]. Hence, this is the required solution.
