Question

The force between two charges when they are 2 cm apart is

0.036 N. If the sum of two charges is 10uC, what are the
charges? (1/4ttɛo=9x109 Nm-C-2).​

Answer 1

`q_1=9.9998\mu C` and
`q_2=0.0002\mu C`

Or

`q_1=0.00016\mu C` and
`q_2=9.99984\mu C`

Step-by-step explanation:

We are given that

Force between two charges=0.036 N=
`36* 10^(-3)N`

Distance between two charges, r=2cm=
`2* 10^(-2)`m

1m=100cm

Sum of two charges=
`10\mu C`

Let one charge=
`q_1=q\mu C=q* 10^(-6)C`

`q_2=(10-q)* 10^(-6) C`

We know that

Electric force between two charges

`F=(kq_1q_2)/(r^2)`

Where
`k=(1)/(4\pi \epsilon_0)=9* 10^(9)`

Using the formula

`36* 10^(-3)=9* 10^(9)* (q* 10^(-6)*(10-q)* 10^(-6))/((2* 10^(-2))^2)`

`(144* 10^(-7))/(9* 10^(9)* 10^(-12))=q(10-q)`

`0.0016=10q-q^2`

`q^2-10q+0.0016=0`

`10000q^2-100000q+16=0`

`q=(100000\pm√((100000)^2-4* 10000* 16))/(2* 10000)`

Using the formula

`x=(-b\pm √(b^2-4ac))/(2a)`

`q=9.999` and
`q=0.00016`

`q_2=10-9.9998=0.0002`

`q_2=10-0.00016=9.99984`

Hence, two charges are

`q_1=9.9998\mu C` and
`q_2=0.0002\mu C`

Or

`q_1=0.00016\mu C` and
`q_2=9.99984\mu C`