- Download PDF
Five equal charges Q = 6.4 x 10^-19C are equally spaced on a semicircle or radius R = 0.10 meters. Find the force on a charge q = 3.2 x 10^-19 C located at the center of the semicircle.
I know it has something to do with the x and y components but I do not understand. Please help.
1 Answer | Add Yours
`F=k(Qq)/(r^2)` where `k ~~8.987551 xx10^9 N*m^2//C^2`
Equally spaced means one a `0^o` , `45^o` , `90^o` , `135^o` , and `180^o`
The forces at `0^o` and `180^o` cancel out, but we can include them anyway.
We will label the forces from `0^o` to `180^o` from 1 to 5.
In all these cases |F| is the same `= k(Qq)/(r^2)` where `Q = 6.4xx10^-19 C` ,
`q = 3.2xx10^-19 C` , and r = 0.1 m we get
`|F| = (8.987551x10^9)((6.4xx10^-19)(3.2xx10^-19))/(0.1)^2 = 1.84 xx 10^-25 N`
`F_1 = |F|haty`
`F_2 = |F|(sqrt(2)/2hatx + sqrt(2)/2 haty)`
`F_3 = |F|hatx`
`F_4 = |F|(sqrt(2)/2hatx - sqrt(2)/2 haty)`
`F_5 = -|F|haty`
Adding these up we get
`F_("tot") = |F|(1+sqrt(2))hatx`
`F_tot = 1.84xx10^-25 * 2.141 = 3.939 xx 10^-25 N`
Our final answer is `3.9 xx 10^-25 N` . This is directed away from the semicircle in the direction of the radius line from the middle of the semicircle to the center.
We’ve answered 319,233 questions. We can answer yours, too.Ask a question