Answer:
E=3 x 10^4 N/c
Explanation:
The electric field strength can be found out disk with a uniform positive surface charge density by
[tex]E= (\sigma/\2epsilon_o)(1-z/ \sqrt(z^2+r^2))[/tex]
σ= charge density
r= radius of the disk
z= position in which we have to find electric field = 15 cm
ε_0= constant ( vacuum permitivity)
putting values we get
[tex]E= \frac{2.5\times10^{-6}}{2\times2.5\times10^{-6}}(1-\frac{0.15}{\sqrt{0.15^2+0.075^2} })[/tex]
solving we get
E=30000 N/c
E=3 x 10^4 N/c
