A charge of 2 μC is located 48 cm above the origin. A second charge of -8 μC is located 48 cm below the origin. What is the magnitude and direction of the net electric field 20cm to the right of the origin?

1 Answer
Michael
Apr 5, 2018

#sf(E=316.7xx10^3color(white)(x)"N/C")# at an angle of #sf(75.9^@)# below the horizontal or bearing #sf(284.1^@)#

Explanation:

The electric field at a particular point is the force experienced by a test charge of +1 Coulomb.

I will assign a charge of +1C at the location specified and find the net force that it experiences. The electric field is found by dividing that force by unit charge.

I will show 2 ways that you can do this.

(1)

Coulombs Law gives us:

#sf(F=k.(q_1q_2)/r^2)#

#sf(k=9.00xx10^(9)color(white)(x)N.m^2"/"C^2)#

Here are the forces acting:

MFDocs

From Pythagoras we get:

#sf(r^2=48^2+20^2)#

#sf(r=52color(white)(x)cm=0.52color(white)(x)m)#

#:.##sf(F_1=(9xx10^9xx2xx10^(-6)xx1)/0.52^2=66.57xx10^(3)color(white)(x)N)#

#sf(F_1=66.57color(white)(x)kN)#

By inspection you can see that #sf(F_2)# must be 4 x this:

#sf(F_2=66.57xx4=266.27color(white)(x)kN)#

Now we find the resultant of these two forces.

We have a S ide A ngle S ide triangle ABC so we can apply The Cosine Rule:

#sf(a^2=b^2+c^2-2"bc"cosA)#

From the diagram you can see that #sf(tantheta=48/20=2.4)#

From which #sf(theta=67.38^@)#

#sf(A=2theta=2xx67.38^@=134.76^@)#

#:.##sf(F_(res)^2=66.57^2+266.27^2-2xx66.57xx266.27cos(134.76))#

#sf(F_(res)^2=4,431.5+70,899.7-(-24,962.5))#

#sf(F_(res)=316.69color(white)(x)kN=316.69xx10^3color(white)(x)N)#

#:.##sf(color(red)(E=(316.69xx10^3)/(1)color(white)(x)"N/C")#

To find angle C we can use The Sine Rule:

#sf(a/sinA=c/sinC)#

#:.##sf(316.69/(sin134.76^@)=266.7/sinC)#

#sf(sinC=266.27/446.0=0.597)#

#sf(C=36.65^@)#

#sf(36.65^@+134.76^@+B=180^@)#

#sf(B=8.58^@)#

The angle the resultant makes with the horizontal is therefore #sf(67.38^@+8.56^@=color(red)(75.94^@)#

(2)

We can resolve #sf(F_1)# and #sf(F_2)# into their horizontal and vertical components which can be added and resolved using Pythagoras.

#sf(F_x=F_1costheta-F_2costheta)#

#sf(F_1=66.57xx0.3846-266.27xx0.3846color(white)(x)kN)#

#sf(F_x=25.603-102.407=76.80color(white)(x)kN)#

#sf(F_y=F_1sintheta+F_2sintheta)#

#sf(F_y=66.57xx0.92307+266.27xx0.92307color(white)(x)kN)#

#sf(F_y=61.449+245.785=307.23color(white)(x)kN)#

Now we can apply Pythagoras:

MFDocs

#sf(F_(res)^2=76.80^2+307.23^2)#

#sf(F_(res)=sqrt(100,288.5)=316.68color(white)(x)kN)#

#:.##sf(color(red)(E=(316.68xx10^3)/(1)color(white)(x)"N/C")#

#sf(tanalpha=307.23^@/76.8^@=4)#

#sf(color(red)(alpha=75.96^@)#

As you can see there is close agreement between the 2 methods so thats all good.

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