Acharge q= 2 c is distributed uniformly through out a bar of length l=2.5 m. the bar is placed horizontally in free space. a second charge q = 10−9c is placed along the line of the bar a distance d= 2m away in space, measured from the right end of the bar. what force is exerted on charge q by the charged bar?

F = 2 N

Explanation:

Let the rod is made up of large number of point charges

so here force due to one small part which is at distance "x" from the end of the rod on the given charge is

here we know that

so we have

now plug in all data

  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