Question

The size of the electric field force in A = 150 N/C and in B = 100 N/C. What's the size of the charges Q1 and Q2? Thank you!

Answer 1

This is what I get

Explanation:

We know that Electric field `vecE` is defined as the electric force produced at a distance `R`, per unit charge. The direction of the electric field is the direction of the force charge `Q` would exert on a positive test charge.

`vecE=(kQ)/R^2`
where for air `k~~9.0 xx 10^9 Nm^2C^-2`

Let direction `Q_1->Q_2` be positive `x`-axis.

Both Charges are shown as positive. Setting up the equation for electric field at `A` using SI units

`vecE_A=[(kQ_1)/(0.1)^2-(kQ_2)/(0.2)^2]hati=150hat iNC^-1`

Multiplying both sides with `(0.2)^2` to remove fraction from denominators and equating magnitudes we get

`(0.2)^2[(kQ_1)/(0.1)^2-(kQ_2)/(0.2)^2=150]`
`=>k(4Q_1-Q_2)=6`
`=>4Q_1-Q_2=6/k` .......(1)

Similarly for point `B`

`vecE_B=[(kQ_1)/(0.2)^2-(kQ_2)/(0.1)^2]hati=-100hat iNC^-1`

Multiplying both sides with `(0.2)^2` to remove fraction from denominators and equating magnitudes we get

`(0.2)^2[(kQ_1)/(0.2)^2-(kQ_2)/(0.1)^2=-100]`
`=>k(Q_1-4Q_2)=-4`
`=>Q_1-4Q_2=-4/k` .......(2)

Multiplying (2) with `4` and subtracting from (1) we get

`4Q_1-Q_2-4xx(Q_1-4Q_2)=6/k-(4xx(-4/k))`
`=>15Q_2=22/k`
`=>Q_2=22/(15k)`

Inserting value of `k`

`=>Q_2=22/(15xx9xx10^9)=1.63xx10^-10C`, rounded to two decimal places.

from (2)

`Q_1=4Q_2-4/k`

Inserting value of `Q_2` we get

`Q_1=4xx(22/(15k))-4/k`
`=>Q_1=88/(15k)-4/k`
`=>Q_1=28/(15k)`

Inserting value of `k`

`Q_1=28/(15xx9xx10^9)`
`=>Q_1=2.07xx10^-10C`, rounded to two decimal places.