Two charges 15nC and 20nC placed at vertex of an equilateral triangle of side 5cm.What is the electric field at the third vertex?tin field of 1.4T. Length of conductor
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Two charges 15nC and 20nC placed at vertex of an equilateral triangle of side 5cm.What is the electric field at the third vertex?tin field of 1.4T. Length of conductor Explanation:
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Given Info : Two charges 15 nC and 20 nC placed at vertex of an equilateral triangle of side 5 cm.
To find : the electric field at the 3rd vertex is...
solution : let ABC is an equilateral triangle where two identical charges are placed at A and B.
we have to find the electric field at C.
so, electric field due to A , E₁ = kq₁/r²
= (9 × 10^9 × 15 × 10^-9)/(5 × 10^-2)²
= 135 × 10⁴/25
= 5.40 × 10⁴ N/C
electric field due to B, E₂ = kq₂/r²
= (9 × 10^9 × 20 × 10^-9)/(5 × 10^-2)²
= 180 × 10⁴/25
= 7.2 × 10⁴ N/C
angle between them is 60° so net electric field, E = √{E₁² + E₂² + 2E₁E₂cos60°}
= 10⁴√{(5.4)² + (7.2)² + (5.4)(7.2)}
≈ 11 × 10⁴ N/C
= 1.1 × 10⁵ N/C
Therefore the electric field at the third vertex is 1.1 × 10⁵ N/c.