The force exerted by an electric charge at the origin on a charged particle at a point (x, y, z) with position vector r = x, y, z is F(r) = Kr/|r|3 where K is a constant. Find the work done as the particle moves along a straight line from (3, 0, 0) to (3, 2, 5).

Explanation:

The parametric representation of a line segment joining the points (a,b,c) and (l,m,n) is

r(t) = (1-t) . (a,b,c) + t . (l, m, n)  where t ∈ |0, 1|

So, the parametric representation of a line segment joining the points (3,0,0) and (3,2,5) is

r(t) = (1 - t) . (3,0,0) + t . (3,2,5)  where t ∈ |0, 1|

r(t) = (3(1 - t), 0, 0) + (3t, 2t, 5t)  where t ∈ |0, 1|

r(t) = (3, 2t, 5t)

Given:

dr = (0, 2, 5) dt

Substitute = 9 + 29t² = u, 92tdt = du

Limit changes from 0→1 to 9 → 38

On solving this, we get:

Therefore, work done is

  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