Question

Two point charges q₁(√(10) μC) and q₂(-25 μC) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,
[ take 1/(4πε₀) = 9 × 10⁹ Nm² C⁻² ]

(A) (63î - 27î) ×10² (B) (-63î + 27î) ×10²
(C) (81î - 81î) ×10² (D) (-81î + 81î) ×10²

Answer 1

Answer:

C. will be correct answers

Answer 2

The electric field (in V/m) at a point y = 3 m on the y-axis is:

(63i^−27j^​)×10²V/m

Let us consider E₁​ & E₂​ as the values of the electric field due to charges q₁​ and q₂ respectively.

Now, Magnitude of E₂ ​= ​ q₂  / 4π∈r²

E₂​ = (9×10⁹×(25)×10⁻⁶) / (4²+3²) V/m

E₂​ = 9×10³V/m

Now, E₂ ​= 9×10³(cosθ₂​i^−sinθ₂​j^​)

∴tanθ₂​ = 3/4​

Thus, E₂​ ​= 9x20³(4/5​i^−3/5​j^​)

              = (72i^−54j^​)x10²

Now, the Magnitude of E₁ ​= √10 x 10⁻⁶/4π∈​ (1² + 3²)

                                           = (9×10⁹) x√10  x 10⁻⁷

                                           = 9√10x 10²

Thus,  E₁  ​=9√10x10²[cosθ₁​(−i^)+sinθ₁ ​j^​]

∴tanθ₁ = 3

Thus,  E₁ ​=9√10x 10²[​(−i^) / √10+ 3​(j^) /√10 ​]

          E₁ ​= 9×10² [−i^+3j^​]

              = [−9i^+27j^​]10²

Therefore, E =  E₁ ​+  E₂​

                    ​= (63i^−27j^​)×10² V/m