Question #82085

2 Answers
Nov 2, 2016

i46.tinypic
Let charges be #q_1=+10muC#, #q_2=+8muC# and #q_3=+15muC# be placed as shown in the figure above. All three vertex angles #=60^@# and all sides #a=10cm#
Now according to Coulomb's Law
#vecF_13 = k_e(q_1q_3)/r_13^ 2 hatr_13#

Inserting given values we get
#|vecF_13| = (9.0 xx 10^9)((10xx10^-6)(15xx10^-6))/(0.1)^2=135N#

Similarly
#vecF_23 = k_e(q_2q_3)/r_23^ 2 hat r_23#
From the figure we see that #hat r_23=hati#
Inserting given values we get
#|vecF_23| = (9.0 xx 10^9)((8xx10^-6)(15xx10^-6))/(0.1)^2=108N#

From the figure it is clear that angle between the two unit vectors #hatr_13 and hati# is #60^@#
As such the #x and y# components of #vecF_13# can be written as
#vecF_13=135cos60^@hati-135sin60^@hatj#
#vecF_13=67.5hati-116.9hatj#
Total force is sum of two forces
#vecF_13+vecF_23=67.5hati-116.9hatj+108hati#
#vecF_"total"=(175.5hati-116.9hatj)N#

Nov 9, 2016

Answer:

#sf(211color(white)(x)N)# at an angle of #sf(56^@)# to the vertical.

Explanation:

I will suggest two methods you could use which involves finding the resultant of the two forces acting on the charge.

Here is the situation:

MFDocs

#sf(q_1=+10muC)#

#sf(q_2=+8muC)#

#sf(q_3=+15muC)#

Method (1)

Here is the vector diagram showing the forces acting on #sf(q_3)#:

MFDocs

There is a horizontal force #sf(stackrel(rarr)(F_(23))#. There is a force #sf(stackrel(rarr)(F_(13))# at an angle of #sf(60^@)# to the horizontal.

These are indicated by the red arrows. We need to get the resultant R.

Coulombs Law gives us the force of attraction between #sf(q_2)# and #sf(q_3)#:

#sf(stackrel(rarr)(F_(23))=(kq_2q_3)/(r_(23)^2).hatr)#

#sf(k)# is a constant with the value #sf(9xx10^(9)" "N.m^(2).C^(-2))#

#:.##sf(stackrel(rarr)(F_(23))=(kxx8xx10^(-6)xx15xx10^(-6))/(0.1^2)=120kxx10^(-10)" "N)#

By the same reasoning:

#sf(stackrel(rarr)(F_(13))=(kxx10xx10^(-6)xx15xx10^(-6))/(0.1^2)=150kxx10^(-10)" "N)#

To make the numbers easier to handle I will normalise the forces in units of #sf(K)# where #sf(K=kxx10^(-10)" "N)#

Now we have a Side Angle Side triangle. We know 2 sides and the angle between them so we can apply the cosine rule to find the unknown side.

For a triangle abc this gives us:

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

Applying this to the vector diagram we get:

#sf(R^2=150^2+120^2-(2xx150xx120xxcos120)#

#sf(R^(2)=22500+14400-(-18000))#

#sf(R^2=54900)#

#sf(R=sqrt(54900)=234.3" "K)# units

Re-scaling:

#sf(R=234.3xx9xx10^9xx10^(-10)=color(red)(211)" "N)#

To find the angle #sf(theta)# we can apply the sine rule:

#sf(234/sin120=120/sintheta)#

#:.##sf(sintheta=0.444)#

From which:

#sf(theta=26.3^@)#

From the diagram you can see that resultant R must be at an angle of #sf(30^@+26.3^@=56.3^@)# to the vertical.

Method (2)

We can resolve the forces into their horizontal and vertical components and find the resultant from that.

Horizontal components:

The horizontal component of #sf(stackrel(rarr)(F_(13))=sf(stackrel(rarr)(F_(13))cos60)=stackrel(rarr)(F_(13))xx0.5)#

So the total horizontal component is given by:

#sf(F_x=stackrel(rarr)(F_(23))+0.5xxstackrel(rarr)(F_(13)))#

#sf(F_x=120K+0.5x150K=195K)#

Vertical components:

#sf(stackrel(rarr)(F_(23))# does not have a vertical component so:

#sf(F_(y)=stackrel(rarr)(F_(13))cos30=150Kxx0.866=129.9K)#

Now we can apply Pythagoras:

MFDocs

#sf((129.9K)^2+(195K)^2=R^2)#

#sf(R^(2)=54899K^(2))#

#sf(R=234.3K" "N)#

#sf(R=234.3xx9xx10^(9)xx10^(-10)=color(red)211" "N)#

To find the angle #alpha# to the vertical:

#sf(Tanalpha=195/129.9=1.5)#

From which:

#alpha=56.3^@#

So the 2 methods are in agreement.