Question

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?

Answer 1

`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:

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:

`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.