Two charged particles, with charges q1=qq1=q and q2=4qq2=4q, are located on the x axis separated by a distance of 2.00cm2.00cm . A third charged particle, with charge q3=qq3=q, is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3.Find the position of charge 3 when qqq = 2.00 nCnC .

Charge 3 is located on the x-axis a distance of 0.67 cm from charge 1 and 1.33 cm from charge 2.

Explanation: Electrostatic Force is the force of repulsion or attraction between two charged particles. It's directly proportional to the charge of the particles and inversely proportional to the distance between them:

k is an electrostatic constant

For the system of 3 particles, suppose distance from 1 to 3 is x meters, so, distance from 2 to 3 is (0.02-x) meters.

Force will be

Substituting:

Solving quadratic equation using Bhaskara:

x₂ will give a negative value and since distance can't be negative, use x₁

x₁ = 0.0067 m

The position of charge 3 is 0.67 cm from charge 1 and 1.33 cm from charge 2.

  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