Question

A parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates. A 1.1 g plastic bead with a charge of -5.6 nC is suspended between the two plates by the force of the electric field between them. What is the charge on the positive plate?

Answer 1

Answer:

The charge on positive plate is $-1.92 \times 10^{-7}$ C

Explanation:

Given :

Diameter $d = 0.12$ m

Radius $r = 0.06$ m

Mass of bead $= 1.1 \times 10^{-3}$ Kg

Charge of bead $q = -5.6 \times 10^{-9}$ C

Electric field in capacitor is given by,

   $E = \frac{Q}{\epsilon_{o} A}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

For finding electric field in terms of force,

   $E = \frac{F}{q}$

But $F = mg$

$F = 1.1 \times 10^{-3 } \times 9.8$         ( $g = 9.8 \frac{m}{s^{2} }$ )

$F = 10.78 \times 10^{-3}$ N

So electric field, $E = \frac{10.78 \times 10^{-3} }{-5.6 \times 10^{-9} }$

$E =- 1.92 \times 10^{6}$ $\frac{N}{C}$

Now charge on positive plate is,

$Q = E \epsilon _{o} A$

$Q = -1.92 \times 10^{6} \times 8.85 \times 10^{-12} \times \pi (0.06)^{2}$

$Q =- 1.92 \times 10^{-7}$ C

Therefore, the charge on positive plate is $-1.92 \times 10^{-7}$ C