Question

particles of charge q and 3q are placed on the x-axis at x=-40 and x=50, respectively. A third particle of charge q is placed on the x-axis, and the total electric force on this particle is zero. Determine the position of the particle.​

Answer 1

(x ≈ -7.06)

Step-by-step explanation:

Given:

• charge on particle 1,
  `=q`

• charge on particle 2,
  `=3q`

• position of particle 1 on x axis,
  `x=-40`

• position of particle 2 on x axis,
  `x=50`

∴The total distance between the charges 1 & 2 is 90 units.

For a third particle of charge 'q' to be placed on x axis so that it experiences no force, let us place it at a distance of r units from the charge q.

∴
`F_(qq)=F_(3qq)`

`k * (q.q)/(r^2) = k * (q.3q)/((90-r)^2)`

`r^2+90r-4050=0`

`r=32.94\ or\ r=-122.94\ (not\ possible)`

`r\approx32.94` from the chage q at x= -40

i.e. position of third charge will be at (x ≈ -7.06)

Answer 2

x = -7.06 cm

Step-by-step explanation:

The distance of new charge q from origin = x

We know that force between two charge given as

`F=K(q_1q_2)/(r^2)`

Given that the force on the third charge q is zero.

Therefore

`K(q* 3q)/((50-x)^2)=K(q* q)/((40+x)^2)`

`( 3)/((50-x)^2)=(1)/((40+x)^2)`

3(40+x)² = (50-x)²

`50-x = √(3)\ (40+x)`

50 - x = 69.28 + 1.73 x

2.73 x = 50 - 69. 28

x = -7.06 cm

Therefore the third charge on the negative x axis and distance from origin will be 7.06 cm.