Answer:
The charge on positive plate is [tex]-1.92 \times 10^{-7}[/tex] C
Explanation:
Given :
Diameter [tex]d = 0.12[/tex] m
Radius [tex]r = 0.06[/tex] m
Mass of bead [tex]= 1.1 \times 10^{-3}[/tex] Kg
Charge of bead [tex]q = -5.6 \times 10^{-9}[/tex] C
Electric field in capacitor is given by,
[tex]E = \frac{Q}{\epsilon_{o} A}[/tex]
Where [tex]\epsilon _{o} = 8.85 \times 10^{-12}[/tex]
For finding electric field in terms of force,
[tex]E = \frac{F}{q}[/tex]
But [tex]F = mg[/tex]
[tex]F = 1.1 \times 10^{-3 } \times 9.8[/tex] ( [tex]g = 9.8 \frac{m}{s^{2} }[/tex] )
[tex]F = 10.78 \times 10^{-3}[/tex] N
So electric field, [tex]E = \frac{10.78 \times 10^{-3} }{-5.6 \times 10^{-9} }[/tex]
[tex]E =- 1.92 \times 10^{6}[/tex] [tex]\frac{N}{C}[/tex]
Now charge on positive plate is,
[tex]Q = E \epsilon _{o} A[/tex]
[tex]Q = -1.92 \times 10^{6} \times 8.85 \times 10^{-12} \times \pi (0.06)^{2}[/tex]
[tex]Q =- 1.92 \times 10^{-7}[/tex] C
Therefore, the charge on positive plate is [tex]-1.92 \times 10^{-7}[/tex] C
