Answer to Question #215469 in Electric Circuits for Tianye Kwadwo Fred

Question #215469
Four electrical point charges with magnitude +100 form a square with a side of 2 m. find the magnitude and the direction of the net electric force acting at the bottom right of the square.
1
Expert's answer
2021-07-12T02:48:01-0400

FAB=kq1q2rAB2=k×102×10222=k×1044F_{AB}=\frac{kq_1q_2}{r_{AB}^2}=\frac{k\times10^2\times10^2}{2^2}=\frac{k\times10^4}{4}

FCB=kq1q2rCB2=k×102×10222=k×1044F_{CB}=\frac{kq_1q_2}{r_{CB}^2}=\frac{k\times10^2\times10^2}{2^2}=\frac{k\times10^4}{4}


Fnet=Fnet+FDBF_{net}=F'_{net}+F_{DB}

FDB=kq1q2rDB2=k×102×102(22)2=k×1048F_{DB}=\frac{kq_1q_2}{r_{DB}^2}=\frac{k\times10^2\times10^2}{(2\sqrt2)^2}=\frac{k\times10^4}{8}

Fnet=FAB2+FCB2=2FABF'_{net}=\sqrt{F_{AB}^2+F_{CB}^2}=\sqrt{2}F_{AB}

Fnet=2k×1044F'_{net}=\sqrt{2}\frac{k\times10^4}{4}


Fnet=Fnet+FDB=(2+12)k×1044F_{net}=F'_{net}+F_{DB}=(\sqrt{2}+\frac{1}{2})\frac{k\times10^4}{4}

Fnet=(22+1)9×10138NF_{net}=(2\sqrt{2}+1)\frac{9\times10^{13}}{8}N


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Comments

Tianye Kwadwo Fred
30.04.23, 20:06

A very good answer. Thank you very much.

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