Answer: The upper plate is positive charged. b) 0.53 µC.
Explanation:
As the plastic bead is suspended between the two plates, the external forces acting upon it must be equal and opposite each other.
The external forces acting upon it, are gravity (always downward) and the on due to electric Field, which must aim upwards, to keep the system in equilibrium.
As the charge on which the force must be act is a negative charge, this means that the force and the electric field must aim in opposite direction, so the electric field must point from the upper plate to the lower one.
This means that the upper plate must be positive charged.
We can find out the magnitude of the electric field, as follows:
E = mg / q = 0.0014 Kg. 9.8 m/s2 / 4.4 x 10⁻⁹ C = 3.12. 10⁶ V/m
b) In order to know the charge on the positive charged plate, we know that the capacitance , by definition, is as follows:
C = Q /V
It can be showed, that for a parallel plate capacitor, the capacitance can be expressed as a relationship of geometric parameters, as follows:
C = ε₀ A / d
But, V and E are related linearly by the expression:
V= E. d
Equating both expressions for C, we have:
Q / E.d = ε₀ A / d , so we can simplify and solve for Q, as follows:
Q = ε₀ A E = 8.85. 10⁻¹² farad/m. π (0.156)2 . ¼. 3.12. 10⁶ V/m = 0.53 µC
