Question

A parallel-plate capacitor is constructed of two horizontal 16.8-cm-diameter circular plates. A 1.8 g plastic bead, with a charge of -4.4 nC is suspended between the two plates by the force of the electric field between them.
a. Which plate, the upper or the lower, is positively charged?
b. What is the charge on the positive plate?

Answer 1

Answer:

Explanation:

Given that:

diameter of the circular plates = 16.8 cm

mass of the plastic bead = 1.8 g

charge q = -4.4 nC

From above, the area of the circular plates is:

$Area = \pi r^2$

$Area = \pi (\dfrac{d}{2})^2$

$Area = \pi (\dfrac{16.8}{2*100} m)^2$

Area = 0.022 m²

Thus, as the plastic beads glide between the two plates of the capacitor, there exists a  weight acting downwards while the weight is balanced by the force of the plates acting upwards.

Hence, this can be achieved only when the upper plate is positively charged.

b)

Recall that

Force (F) = qE

where;

F = mg

mg = qE

$E = \dfrac{mg}{q}$

$E = \dfrac{1.8*10^{-3}*9.8}{4.4*10^{-9}}$

E = 4.0 × 10⁶ N/C

From the electric field;

$E = \dfrac{\Big(\dfrac{Q}{A}\Big)}{e_o}$

$4.0*10^{6} = \dfrac{\Big(\dfrac{Q}{0.022}\Big)}{8.85*10^{-12}}$

$4.0*10^{6}*8.85*10^{-12} = {\Big(\dfrac{Q}{0.022}\Big)}{}$

$Q= 4.0*10^{6}*8.85*10^{-12}*0.022$

Q = 7.788 × 10⁻⁷ C

Q = 779 nC