Three point charges lie in a straight line along the y-axis. A charge of q1 = -10.00 µC is at y = 6.40 m, and a charge of q2 = -7.70 µC is at y = -3.60 m. The net electric force on the third point charge is zero. Where is this charge located?

y coordinate where field is zero is  1.1 m

Explanation:

given data

charge q1 = -10.00 µC

y = 6.40 m,

charge q2 = -7.70 µC

y = -3.60 m

solution

we know that here Electric field E that is express as

Electric field E =   ..............................1

and

Distance between charges will be

Distance between charges = 6.40 -  (-3.60) = 10 m

so here neutral point P is at  distance d from q1

so it is  distance (10 - d) from q2.  

so here

Field from q1 at P =

and

Field from q2 at P =

so here field is  in opposite directions and the resultant field = 0

so

d =  5.326242 m

so y coordinate where field is zero is 6.40 - 5.32 = 1.1 m

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂