Question

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

Answer 1

Angle with the vertical: $19.8^{\circ}$

Explanation:

To solve this problem, we have to consider the forces acting in the horizontal and vertical direction.

In the horizontal direction, we have:

• The electric force on the sphere, $F_E=qE$, where $q=5.20\cdot 10^{-8}C$ 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 $Tsin \theta$, where T is the tension and $\theta$ the angle with respect to the vertical

In the vertical direction, we have:

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

Since the sphere is in equilibrium, we can write:

$qE=T sin \theta\\mg = T cos \theta$

The electric field produced by a charged sheet is:

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

where $\sigma$ is the charge density on the sheet.

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

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

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