Electric charge can accumulate on an airplane in flight. You mayhave observed needle-shaped metal extensions on the wing tips andtail of an airplane. Their purpose is to allow charge to leak offbefore much of it accumulates. The electric field around the needleis much larger than the field around the body of the airplane and, can become large enough to produce dielectric breakdown of the air, discharging the airplane. To model this process, assume that twocharged spherical conductors are connected by a long conductingwire and a charge of 22.0 µC isplaced on the combination. One sphere, representing the body of theairplane, has a radius of 6.00 cm, and the other, representing thetip of the needle, has a radius of 2.00 cm.(a) What is the electric potential of eachsphere?(b) What is the electric field at the surface of each sphere?

(a) = 24.75x V

(b) V/m

V/m

Explanation: Eletric Potential (V) is the amount of energy necessary to move a charged particle inside an electric field. It is calculated as:

where:

k is coulomb's constant: k = 9. N.m²/C²

q is the charge of the object

r is the distance

Electric Field (E) is what surrounds an electric particle in a way that every particle inside the field is influenciated by it, through force of attraction or of repulsion. When related to electric potential, can be calculated as: E =

a) The ratio of the two charges is proportional to the ratio of the two radii:

where 1 represents the sphere of the body of the airplane and 2 is the tip of the needle.

(1)

The combine charges of spheres results in a charge of 22.0µC, which means:

22.

Substitute and resolve:

= 22.

C

Using (1) to find the other charge:

= 16.5. C

Now, to determine electric potential for each sphere:

Electric Potential for Sphere 1:

24.75. V

Electric Potential for Sphere 2:

24.75. V

The electric potential of each sphere is the same and has magnitude 24.75. V.

b) Electric Field for Sphere 1:

V/m

Electric field for Sphere 2:

= V/m

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

need to see circuit to get time constant