The electric field at (30, 30) cm due to a charge of -8 nC at the origin in NC^-1 is

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$- 200 \sqrt{2} (\hat{i} + \hat{j})$

$200 \sqrt{2} (\hat{i} + \hat{j})$

$400 (\hat{i} + \hat{j})$

$- 400 (\hat{i} + \hat{j})$

answer is C.

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Detailed Solution

We need to calculate the electric field at point (30, 30) cm due to a charge of -8 nC located at the origin.

Step 1: Recall the Electric Field Formula

Electric Field: E = kq/r² × r̂

Step 2: Calculate the Distance r

Given point coordinates: (30 cm, 30 cm) = (0.3 m, 0.3 m)

r = √[(0.3)² + (0.3)²] = √(0.09 + 0.09) = √0.18 ≈ 0.424 m

Step 3: Calculate the Electric Field Magnitude

E = (9 × 10⁹ × 8 × 10⁻⁹) / (0.424)²

E = (72 × 10⁰) / 0.18 = 400 N/C

Step 4: Determine the Direction

The unit vector r̂ in the direction of the point is:

r̂ = (0.3 î + 0.3 ĵ) / 0.424 ≈ (1/√2) (î + ĵ)

Since the charge is negative, the electric field will point toward the charge (opposite to r̂).

E = -400 × (1/√2) × (î + ĵ) = -200√2 × (î + ĵ)

The correct answer is: (c) -200√2 î + ĵ

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