Two point charges 4*10^-6C and 2*10^-6C are placed at the vertices A and B of a right angled triangle ABC respectively. B is the right angle, AC=2*10^-2 m and BC=10^-2m. Find the magnitude and direction of resultant electric intensity at C?

The electric field strength or intensity is the force that a test charge of +1C would experience at that position in the field.

I will place a +1C charge at the location specified and find the resultant force acting on it in N. By dividing this by +1C I will get the electric field strength in N/C.

The forces look like this.

#sf(cosBhatCA=0.01/0.02=0.5)#

#:.##sf(BhatCA=60^@)#

#:.##sf(BhatAC=30^@)#

Coulomb's Law gives:

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

#sf(k)# is a constant and = #sf(9xx10^(9)color(white)(x)"Nm"^2"C"^(-2))#

#:.##sf(F_1=(9xx1xx2xx10^(-6))/0.01^2color(white)(x)N)#

#sf(F_1=1.8xx10^(8)color(white)(x)N)#

#sf(F_1=180color(white)(x)MN)#

#sf(F_2=(9xx10^(9)xx4xx10^(-6)xx1)/(0.02^2)color(white)(x)N)#

#sf(F_2=0.9xx10^(8)color(white)(x)N)#

#sf(F_2=90color(white)(x)MN)#

To find the resultant R of these 2 forces we can apply The Cosine Rule.

This gives:

#sf(R^2=F_1^(2)+F_2^(2)-(2xxF_1xxF_2cos120))#

#sf(R^2=180^(2)+90^(2)-(2xx180xx90cos120))#

#sf(R^2=32,400+8,100-(2xx32,400xx8,100xx-0.5)#

#sf(R^(2)=56,644)#

#sf(R=238color(white)(x)MN=2.38xx10^(8)color(white)(x)N)#

To find the electric field strength we divide by 1 N:

#sf(E=+(2.38xx10^(8))/1=+2.38xx10^(8)color(white)(x)"N""/""C")#

To find the angle #sf(theta)# we can apply The Sine Rule :

#sf((sintheta)/180=sin120/238)#

#sf(sintheta=(0.866xx180)/238=0.6549)#

From which:

#sf(theta=40.9^@)#

From the diagram you can see that:

#sf(theta+delta=60^@)#

#:.##sf(delta=60-theta=60-40.9=19.1^@)#

This is a bearing of #sf(099.1^@)#

So we can say that:

#sf(E=+2.38xx10^(8)color(white)(x) "N/C")# at an angle of #sf(19.1^@)# to the horizontal.