Question

A parallel-plate capacitor is constructed of two horizontal 13.2-cm-diameter circular plates. A 1.5 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. Part A Which plate, the upper or the lower, is positively charged? Upper. Lower. It is impossible to tell. Submit Request Answer Part B What is the charge on the positive plate? Express your answer with the appropriate units. A MO ? Value Units q= Submit Request Answer

Answer 1

Given, diameter of circular plate ( d ) = 13.2 cm =

132 x 10-3m

mass of the plastic bead ( m ) = 1.5 g =

15 x 10-4kg

charge of the plastic bead ( q ) = -4.4 nC =

-44 x 10-100

Assume, acceleration due to gravity ( g ) =

10m/s?

(a) The bead is suspended due to electric field force against the force of grvity.

As the given charge is negative. So, the force due to electric field will be opposite to the direction of electric field.

The force due to gravity will be downwards. So, the force due to electric field will be upward.

Therefore, direction of electric field will be downward.

For the direction to be downward, the upper plate must be positive charged and the lower plate negative charged.

So, the upper plate will be positively charged.

(b) The electric filed between the opposite charged plates of capacitor,

  $E=\frac{\sigma }{\varepsilon_o }$ ......( 1 )

where $\sigma$ is the surface charge density of the plates

  $\varepsilon _o$ is the permittivity of free space =

8.85 x 10-12kg-m-34A

Balancing the force on the bead,

  

FE = mg

  

qE = mg

E = mg

E = (1.5 x 10-3) (10) (4.4 x 10-9)

E = 3.41 x 10V/m

   putting value in equation ( 1 ),

  

o = E.E.

o = (3.41 x 10) x (8.85 x 10-12

  

0 = 3.02 x 10-5C/m

We know,

  $\sigma = \frac{Q}{A}$

where Q is the charge on the surface

A is the area of the surface

Here, we have the circular surface.

  

ο = σXA

Q = (3.02 x 10-5) x (x (6.6 x 10-2)

Q = 3.02 x 3.14 x 6.6 x 6.6 x 10-9

Q = 413.07077 x 10-9

Q = 413.07nc