Question

The electric field at point (30, 30, 0) due to a charge of \(0.008 \, \mu{C}\) placed at the origin will be: (coordinates are in cm)

• A. \(8000 \, {N/C} \, \hat{i} + 8000 \, {N/C} \, \hat{j}\)
• B. \(4000(\hat{i} + \hat{j}) \, {N/C}\)
• C. \(200\sqrt{2}(\hat{i} + \hat{j}) \, {N/C}\)
• D. \(400\sqrt{2}(\hat{i} + \hat{j}) \, {N/C}\)

Hint:

The electric field due to a point charge decreases with the square of the distance and points radially away from the charge.

Answer 1

The Correct Option is C

Step 1: {Calculating the distance}
\[ r = \sqrt{(30)^2 + (30)^2 + 0^2} = 30\sqrt{2} \, {cm} = 30\sqrt{2} \times 10^{-2} \, {m} \] Step 2: {Calculating the electric field}
\[ E = \frac{kq}{r^2} = \frac{9 \times 10^9 \times 0.008 \times 10^{-6}}{(30\sqrt{2} \times 10^{-2})^2} = \frac{72}{18} \times 10^3 = 4000 \, {N/C} \] Step 3: {Direction of the electric field}
The electric field components in the \(x\) and \(y\) directions are equal: \[ E_x = E_y = \frac{4000}{\sqrt{2}} = 200\sqrt{2} \, {N/C} \] Thus, the electric field is: \[ \vec{E} = 200\sqrt{2}(\hat{i} + \hat{j}) \, {N/C} \]