The electric field is calculated as [tex]3.09*10^8N/C[/tex]
Data;
- Mass = 8.8mg
- charge (q) = 72.0nC
- length of strings = 0.530m
- angle = 58 degrees
The Electric Field
Let the distance between the two charge be represented by x.
The force on the charge due to the other is calculated as
[tex]F= \frac{kq^2}{x^2}[/tex]
The distance between the two charges will be
[tex]sin\frac{\theta}{2} = \frac{x}{2L} \\x = 2L sin\theta /2[/tex]
In the vertical direction, the forces are balanced by
[tex]Tcos \frac{\theta}{2} =mg\\ T = \frac{mg}{cos \theta / 2}[/tex]
Balancing the forces in the horizontal direction;
[tex]qE-Tsin\frac{\theta}{2} - F =0\\ qE - Tsin\frac{\theta}{2}- \frac{kq^2}{x^2}= 0\\ qE = Tsin\frac{\theta}{2}+ \frac{kq^2}{x^2} \\ qE = \frac{mgsin\frac{\theta}{2} }{cos\frac{\theta}{2} } + \frac{kq^2}{(2L sin\frac{\theta}{2})^2 } \\E = \frac{mgtan\frac{\theta}{2} }{q} + \frac{kq^2}{(2Lsin\frac{\theta}{2})^2 }[/tex]
The electric field will be
[tex]E = \frac{8.80*10^-^6*9.8*tan(\theta /2)}{72*10^-^9} + \frac{9*10^9*72*10^-^9}{2*(0.5330)sin(58/2)}\\ E = 3.09*10^8N/C[/tex]
The electric field is calculated as [tex]3.09*10^8N/C[/tex]
Learn more on electric field around a sphere here;
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