Question

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?

Answer 1

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

Explanation:

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.