Question

ABCD is a square of side √2m. Point charges of 2nC, 3nC 4nC and 5nC are placed at corners A, B, C and D respectively. Calculate the electric intensity at point O. please answer this question with steps..​

Answer 1

Given info : ABCD is a square of side √2m. Point charges of 2nC, 3nC 4nC and 5nC are placed at corners A, B, C and D respectively.

To find : electric field intensity at point O ( intersecting point of diagonals ) is...

solution :

Electric field due to A at O , $E_A$ = k(2nC)/(AO)²

Electric field due to B at O, $E_B$ = K(3nC)/(BO)²

electric field due to C at O, $E_C$ = k(4nC)/(CO)²

electric field due to D at O, $E_D$ = k(5nC)/(DO)²

here, AO = BO = CO = DO = diagonal/2

= √2 × side length/2 = side length/√2 = √2/√2 = 1 m

see figure,

it is clear that,$E_A$ is acting just opposite of $E_C$.

similarly, $E_B$ is acting just opposite of $E_A$.

here, $E_A$ < $E_C$ and $E_B$ < $E_D$

then, Electric field along OA = $E_C-E_A$ = k(4nC)/1² - k(2nC)/1² = k(2nC) = E (let)

electric field along OB = $E_D-E_B$ = k(5nC)/1² - k(3nC)/1² = k(2nC) = E

now you see, angle between OA and OB is 90°

so resultant electric field, E = √(E² + E²)

= √2E

= √2k(2nC)

= √2 × 9 × 10^9 × 2 × 10^-9 N/C

= 18√2 N/C

Therefore the electric field intensity at O due to all four charges is 18√2 N/C.

Answer 2

electric charges and field numerical