Question

A proton is on the x axis at x = 1.2 nm . An electron is on the y axis at y = 0.80 nm . Part A Find the net force the two exert on a helium nucleus (charge +2e) at the origin.

Answer 1

The net force on the helium nucleus at the origin is equal to
`78.79* 10^(-11)`

Step-by-step explanation:

Here let the force exerted by the proton be
`F_(1)` and force exerted by electron be
`F_(2)`.

⇒The net force becomes
`\sqrt{F1^(2)+F2^(2) }`

Now lets calculate
`F_(1)` and
`F_(2)`

F =
`(KQq)/(r^(2) )` ; where for
`F_(1)` r=
`1.2nm` and for
`F_(2)` r=
`O.8nm`

upon calculating we get
`F_(1)`=
`32* 10^(-11)` and
`F_(2)`=
`72* 10^(-11)`

⇒F =
`78.79* 10^(-11)`