Question

Three point charges are placed at the vertices of a right isosceles triangle as
shown in the Fig. a. What is the magnitude and
direction of the resultant electric field at point P which is the mid point of its​

Answer 1

hypotenuse= 7.07106781186

Answer 2

Given info : Three point charges are placed at the vertices of a right isosceles triangle as shown in figure.

To find : the magnitude and the direction of the resultant electric field at point P which is the midpoint of its, are ...

solution : BP = √(BA² - PA)² = √{BA² - (AC/2)²}

= √{5² - (5/√2)²}

= 5/√2

electric field due to B, $E_B$ = k(10μC)/(5cm/√2)²

= (9 × 10^9 × 10^-5)/(25/2 × 10^-4)

= 18 × 10^8/25

= 72 × 10^6 N

in vector form,

$E_B$ = 72 × 10^6 (cos45° i + sin45° j)

= 36√2 × 10^6(i + j)

similarly, electric field due to A, $E_A$ = k(10μC)/(5/√2 cm)² = 72 × 10^6 N

in vector form,

$E_A$ = 72 × 10^6(cos45° i - sin45° j)

= 36√2 × 10^6(i - j)

electric field due to C, $E_C$ = 72 × 10^6 N

in vector form,

$E_C$ = 72 × 10^6(-cos45i + sin45° j)

= 36√2 × 10^6 (-i + j)

now net electric field at the point P, E = $E_A+E_B+E_C$

= 36√2 × 10^6 [(i - j) + (i + j) + (-i + j)]

= 36√2 × 10^6(i + j)

Therefore the magnitude of electric field, 36√2 × 10^6 √(1² + 1²) = 72 × 10^6 N

and the direction of electric field, tanθ = y/x = 1/1 = tan45°

so, θ = 45° means resultant electric field makes an angle 45° with x - axis.