Question

Two point charge of 100 mC and -4 mC are 5. positioned at points (-$2\sqrt{3} , 3\sqrt{3} ,-4$) and ($4\sqrt{3} ,-5\sqrt{3} ,6$) respectively of a Cartesian coordinate system. Find the force vector on the -4 mC charge?
All the coordinates are in meters.(Please explain your answer)
(1) 9 x 10^-4 (3√3i - 4√3j + 5k)

(2) 9 x 10^-4 (-3√3i+ 4√3j - 5k)

(3) 2.25 x 10^-4 (-3√3i + 4√3j - 5k)

(4)2.25 x 10^-4(3√3i - 4√3j +5k)

Answer 1

√(105/2) square units is answer

Let OA = 2i - j + k

OB = i - 3j - 5k

OC = 3i - 4j - 4k

and AB, BC and CA represent the sides of trinalge ABC

Now, AB = (1 - 2)i + (-3 - 1)j + (-5 - 1)k = -i - 2j - 6k

and |AB| = √{(-1)2 + (-2)2 + (-6)2 } = √(1 + 4 + 36) = √41

BC = (3 - 1)i + (-4 + 3)j + (-4 + 5)k = 2i - j + k

Now, |BC| = √{(2)2 + (-1)2 + 12 } = √(4 + 1 + 1) = √6

and CA = (2 - 3)i + (-1 + 4)j + (1 + 4)k = -i + 3j + 5k

Now, |CA| = √{(-1)2 + 32 + 52 } = √(1 + 9 + 25) = √35

Again, |BC|2 + |CA|2 = 6 + 35 = 41 = |AB|2

So, triangle ABC is a right angle triangle.

Hence, area of triangle ABC = (1/2) * base * altitude

|BC| * |CA|

= (1/2) * √6 * √35

* √(6 * 35)

= (1/2) * √(210)

√(105/2) square units