Respuesta :
Answer:
[tex]Q = 2.8 \times 10^{-7} C[/tex]
Explanation:
If the bead is suspended between the plates, then the electrostatic force applied by the plates is equal to its weight.
mg = Eq
The relation between the electric field and charge in a parallel plate capacitor is
[tex]E = \frac{Q}{\epsilon_0 A} = \frac{mg}{q}\\Q = \frac{mg\epsilon_0 A}{q} = \frac{(1.1\times 10^{-3})(9.8)(8.8\times 10^{-12})(191\times 10^{-4})}{6.4\times 10^{-9}} = 2.8\times 10^{-7}[/tex]
The charge on the positive (upper) plate is [tex]Q = 2.8\times 10^{-7}C[/tex]
- The calculation is as follows;
In the case when the bead is suspended between the plates, so the electrostatic force should be applied by the plates is equalivalent to its weight.
[tex]= \frac{(1.1\times 10^{-3})(9.8)(8.8\times 10^{-12})(191\times 10^{-4})}{6.4\times 10^{-7}} [/tex]
[tex]Q = 2.8\times 10^{-7}C[/tex]
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