Question

Three positively charged particles, with charges q1=q, q2=2q, and q3=q (where q>0), are located at the corners of a square with sides of length d. The charge q2 is located diagonally from the remaining (empty) corner. Find the magnitude of the resultant electric field E⃗ net in the empty corner of the square.

Answer 1

Answer:

$E=K\dfrac{q}{d^2}\left ( (1+\dfrac{1}{\sqrt2})i+(1+\dfrac{1}{\sqrt2})j \right )$

Explanation:

Given that

q₁ , q₂ , q₃

Electric field due to q₁ is E₁

Electric field due to q₂ is E₂

Electric field due to q₃is E₃

θ = 45 °

E₂ have two component

x -direction E₂ cos 45 °

y-direction E₂ sin 45 °

The resultant E

E = (E₃ + E₂ cos 45 °) i + (E₁ + E₂ sin 45 °) j

$E_1=K\dfrac{q_1}{d^2}$

$E_2=K\dfrac{q_1}{2d^2}$

$E_3=K\dfrac{q_1}{d^2}$

Given that

q₁ = q₂ = q₃ = q

$E=K\dfrac{q}{d^2}\left ( (1+\dfrac{1}{\sqrt2})i+(1+\dfrac{1}{\sqrt2})j \right )$