Point charges of + 6 nC, - 6 nC and + 16 nC are placed at the corners A, C and D of a square ABCD of side 0.4 m. Find the electric field strength at B.
Answers
Given : Point charges of + 6 nC, - 6 nC and + 16 nC are placed at the corners A, C and D of a square ABCD of side 0.4 m.
To find : the electric field strength at B.
solution : see figure, E₁ is the electric field strength due to the charge placed at A.
so, E₁ = kq/r² = (9 × 10^9 × 6 × 10^-9)/(0.4)²
= 54/0.16
= 337.5 N/C [ along positive x - axis ]
E₂ is the electric field due to the charge placed at C.
so, E₂ = kq/r² = (9 × 10^9 × 6 × 10^-9)/(0.4)²
= 54/0.16 = 337.5 N/C [ along negative y- axis ]
E₃ is the electric field due to the charge placed at D.
so, E₃ = kq/r² = (9 × 10^9 × 16 × 10^-9)/(0.4√2)²
= (9 × 16)/(0.16 × 2)
= 450 N [ along N - E ]
net electric field strength at B , E = E₁ i - E₂ j + E₃(cos45° i + sin45° j)
= 337.5i - 337.5 j + 450(1/√2 i + 1/√2 j)
= (337.5 + 318.246)i + (-337.5 + 318.246)j
= 655.746i - 19.254j
so, magnitude of electric field strength is 656.03 N
