A small sphere with a mass of 2.00×10−3
g
g
and carrying a charge of 5.20×10−8
C
C
hangs from a thread near a very large, charged insulating sheet, as shown in the figure (Figure 1). The charge density on the sheet is −2.40×10−9
C/
m
2

The electric force on the sphere, [tex]F_E=qE[/tex], where [tex]q=5.20\cdot 10^{-8}C[/tex] is the charge of the sphere and E the magnitude of the electric field
The horizontal component of the tension in the thread, which is [tex]Tsin \theta[/tex], where T is the tension and [tex]\theta[/tex] the angle with respect to the vertical

In the vertical direction, we have:

The weight of the sphere, downward, of magnitude [tex]mg[/tex], where [tex]m=2.00\cdot 10^{-3} g=2.00\cdot 10^{-6} kg[/tex] is the mass and [tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
The vertical component of the tension, upward, of magnitude [tex]Tcos \theta[/tex]

Since the sphere is in equilibrium, we can write:

[tex]qE=T sin \theta\\mg = T cos \theta[/tex]

The electric field produced by a charged sheet is:

[tex]E=\frac{\sigma}{2\epsilon_0}=\frac{2.40\cdot 10^{-9}}{2(8.85\cdot 10^{-12})}=135.6 N/C[/tex]

where [tex]\sigma[/tex] is the charge density on the sheet.

By dividing eq.(1) by eq.(2), we can find the angle:

[tex]tan \theta = \frac{qE}{mg}=\frac{(5.20\cdot 10^{-8})(135.6)}{(2.00\cdot 10^{-6})(9.8)}=0.360\\\theta=tan^{-1}(0.360)=19.8^{\circ}[/tex]

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