Abead with a mass of 0.050 g and a charge of 20 nc is free to slide on a vertical rod. at the base of the rod is a fixed 20 nc charge. in equilibrium, at what height above the fixed charge does the bead rest?

height above the fixed charge is 8.571 cm

Explanation:

given data

mass m = 0.050 g = 5 × kg

charge bead q1 = 20 nC = 20 × C

charge base q2 = 20 nC = 20× C

to find out

what height above the fixed charge does the bead rest

solution

we know that when charge at rest then downward gravitational force is balance by electrostatic force so

  .............1

here k is 9 × Nm²/C² and g = 9.8 m/s² and r is height of bread  

put here all value in equation 1

r² = 7.3469 ×

r = 0.08571 m = 8.571 cm

so height above the fixed charge is 8.571 cm

  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