Answer:
[tex]a=11.86*10^1^5 m/s[/tex]
[tex]a=3.91864074*10^-^2^4 m/s^2[/tex]
[tex]a=3.95426275*10^-^2^4m/s^2[/tex]
Explanation:
From the question we are told that
The surface charge density on the disk is [tex]\sigma=+4.00 \mu C/m^2[/tex]
[tex]\varepsilon _0= 8.85 * 10^–12 F/m.[/tex]
a)Generally the electric field in this situation is given as is mathematically given as
[tex]E=\frac{\sigma}{\varepsilon_0} (1-\frac{R}{\sqrt{2}R}[/tex]
[tex]E=\frac{\sigma}{\varepsilon_0} (1-\frac{1}{\sqrt{2}})[/tex]
[tex]E=67.48*10^3 f/a[/tex]
Generally the Force of release of the electron is mathematically given as
[tex]F=eE[/tex]
[tex]F=ma[/tex]
[tex]a*m=\frac{eE}{1}[/tex]
[tex]a=\frac{1.6*10^-^1^9 *67.48*10^3}{9.1*10^3^1}[/tex]
[tex]a=11.86*10^1^5 m/s[/tex]
b)Generally the electric field in R/100 from the center of the disk is given as is mathematically given as
Generally
[tex]E=\frac{\sigma}{\alpha \varepsilon_0}(1-\frac{1}{\sqrt{(100)^2} }[/tex]
[tex]E=\frac{4.0*10^-^6}{2*8.854*10^-^1^2}(1-\frac{1}{\sqrt{(1+(100)^2} })[/tex]
[tex]E=2.22872692*10^2^7[/tex]
Generally the Force of release of the electron is mathematically given as
[tex]a*m=\frac{eE}{1}[/tex]
[tex]a=\frac{1.6*10^-^1^9 *2.22872692*10^2^7}{9.1*10^3^1}[/tex]
[tex]a=3.91864074*10^-^2^4 m/s^2[/tex]
c)Generally the electric field in R/1000 from the center of the disk is given as is mathematically given as
[tex]E=\frac{\sigma}{\alpha \varepsilon_0}(1-\frac{1}{\sqrt{(1000)^2} }[/tex]
[tex]E=\frac{4.0*10^-^6}{2*8.854*10^-^1^2}(1-\frac{1}{\sqrt{(1+(1000)^2} })[/tex]
[tex]E=2.24898694*10^2^7[/tex]
Generally the Force of release of the electron is mathematically given as
[tex]a*m=\frac{eE}{1}[/tex]
[tex]a=\frac{1.6*10^-^1^9 *2.24898694*10^2^7}{9.1*10^3^1}[/tex]
[tex]a=3.95426275*10^-^2^4m/s^2[/tex]