Two electric charges of +10-9C and -10-9C are placed at the corners A and B of an .equilateral triangle ABC side 5cm.The electric intensity at C

Question

Two electric charges of +10-9C and -10-9C are placed at the corners A and B of an .equilateral triangle ABC side 5cm.The electric intensity at C (A) 1800N/C (B) 3600 N/C (C) 900 N/C (D) 2700 N/C

Tardigrade Admin · Accepted Answer

Net vecE in y direction gets cancelled Field in x direction is given by: E x =(( k × 10-9/(5 × 10-2)2) × cos 600) × 2 =(9 × 109 × 10-9/25 × 10-4) × (1/2) × 2 E x =3600 N / C <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0deac5a6bea14d76fdfb4e3531c2f2b6-.png" />

Q. Two electric charges of $+ 1 0^{- 9} C$ and $- 1 0^{- 9} C$ are placed at the corners A and B of an .equilateral triangle ABC side 5cm.The electric intensity at C

1800N/C

3600 N/C

900 N/C

2700 N/C

Net $E$ in $y$ direction gets cancelled
Field in $x$ direction is given by:
$E_{x} = (\frac{k \times 1 0 ^{- 9}}{( 5 \times 1 0 ^{- 2} ) ^{2}} \times cos 6 0^{0}) \times 2$
$= \frac{9 \times 1 0 ^{9} \times 1 0 ^{- 9}}{25 \times 1 0 ^{- 4}} \times \frac{1}{2} \times 2$
$E_{x} = 3600 N / C$

Q. Two electric charges of +10−9C and −10−9C are placed at the corners A and B of an .equilateral triangle ABC side 5cm.The electric intensity at C