How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes

7.54 x C

Explanation:

The complete question is:

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular aluminum plates, each 19 cm in diameter, separated by 1.0 cm.How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume that value of 3×106N/C of the field causes a spark.

SOLUTION:

As you know that, parallel plate capacitance can be defined as

C=εo A/d

Where,

εo is the permittivity of free space constant, A is area of the capacitor plates, and d is the distance between them

Also, C= Q/V

where,

'Q' defines charge stored on the capacitor and 'V' express the potential difference between the plates.

Equating the equations.

εo A/d =  Q/V

Q= (εo A V)/d

for a uniform electric field (such as the one between the plates of a parallel-plate capacitor)

V= Ed ->where 'E' defines magnitude of the electric field.

Therefore,

Q= (εo A E d)/d  =>εo A E ->eq(1)

The area of the plates is given by

A= πr² => π(d/2)² =>π(0.19/2)

A=0.0284m²

Given: E= 3 x N/C

Substituting all the required values in eq(1)

(1)=> Q= (8.85x) (-.0284)(3 x )

Q= 7.54 x C

Therefore,7.54 x C can be added to each of the plates before a spark jumps between the two plates

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.