Answer to Question #206241 in Electric Circuits for Ywas

Question #206241

Draw the net electric force on a +1-C charge placed at the center of: 

a. A line segment of length 1.0 m with a +1-C charge and a –1-C charge at its endpoints; 

b. An equilateral triangle of side length 1.0 m with alternating +1-C, –1-C, and +1-C charges 

placed at its vertices;

c. A square of side length 1.0 m with alternating +1-C, –1-C, +1-C, and –1-C charges placed 

at its vertices.

d. Calculate the magnitude and determine the direction of the net electric force on the +1-

C charge placed at the center of configurations in a, b, and c.


1
Expert's answer
2021-06-14T15:04:44-0400

Gives

Part

Mid point P(+1C)

Point A(+1C) and B(-1C )

Electric force at centre

FAp=kq1q2r2F_{Ap}=\frac{kq_1q_2}{r^2}

Put value

FAP=9×109×1×10.52=36×109NF_{AP}=\frac{9\times10^9\times1\times1}{0.5^2}=36\times10^9N

FBP=9×109×1×10.52=36×109NF_{BP}=-\frac{9\times10^9\times1\times1}{0.5^2}=-36\times10^9N

Fnet=36×10936×109=0NF_{net}=36\times10^9-36\times10^9=0N

Part (b)

PointA (+1C) point B(-1C) pointC(+1c)

FAP=kq1q2rAP2F_{AP}=\frac{kq_1q_2}{r^2_{AP}}

Put value

FAP=9×109×1×1(13)2=27NF_{AP}=\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt3})^2}=27N

FBP=9×109×1×1(13)2=27NF_{BP}=-\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt3})^2}=-27N

FCP=9×109×1×1(13)2=27NF_{CP}=\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt3})^2}=27N

Fnet=FAP+FBP+FCPF_{net}=F_{AP}+F_{BP}+F_{CP}

Fnet=2727+27=27NF_{net}=27-27+27=27N

Part(c)

Centre of force

rAP=rBP=rCP=rDP=12r_{AP}=r_{BP}=r_{CP}=r_{DP}=\frac{1}{\sqrt2}

FAP=KQ1Q2rA2pF_{AP}=\frac{KQ_1Q_2}{r^2_Ap}

Put value

FAP=9×109×1×1(12)2=18NF_{AP}=\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt2})^2}=18N

FBP=9×109×1×1(12)2=18NF_{BP}=-\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt2})^2}=-18N

FCP=9×109×1×1(12)2=18NF_{CP}=\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt2})^2}=18N

FDP=9×109×1×1(12)2=18NF_{DP}=-\frac{9\times10^9\times1\times1}{(\frac{1}{\sqrt2})^2}=-18N


Fnet=FAP+FBP+FCP+FDPF_{net}=F_{AP}+F_{BP}+F_{CP}+F_{DP}

Put value

Fnet=0NF_{net}=0N

Part(d)

(a)

Due to centre for Line

Fnet=0N

Due to centre of triangle

Fnet=27N

Due to square at centre

Fnet =0N


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