Electromagnetism

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.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes