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: \[ \vec{E} = 200\sqrt{2}(\hat{i} + \hat{j}) \, {N/C} \]

Answer 2

Step 1: Convert units to SI
Position of point P is (30 cm, 30 cm, 0) = (0.3 m, 0.3 m, 0)
Charge \( q = 0.008\, \mu C = 0.008 \times 10^{-6} \, C \)
\[ \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, Nm^2/C^2 \]

Step 2: Find displacement vector from origin to point P
\[ \vec{r} = 0.3 \hat{i} + 0.3 \hat{j} \quad \text{(in meters)} \]
\[ |\vec{r}| = \sqrt{(0.3)^2 + (0.3)^2} = \sqrt{0.18} = \sqrt{18} \times 0.1 = 0.424 \, m \]

Step 3: Electric field due to point charge
\[ \vec{E} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r^2} \cdot \hat{r} \]
First, calculate \( r^2 = 0.18 \)
\[ E = \frac{9 \times 10^9 \cdot 0.008 \times 10^{-6}}{0.18} = \frac{72 \times 10^3}{0.18} = 4 \times 10^5 \, N/C \]

Step 4: Direction of electric field
Since the displacement vector is equal in x and y, the unit vector in that direction is:
\[ \hat{r} = \frac{0.3 \hat{i} + 0.3 \hat{j}}{\sqrt{0.18}} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{j}) \]

Step 5: Final electric field vector
\[ \vec{E} = 4 \times 10^5 \cdot \frac{1}{\sqrt{2}}(\hat{i} + \hat{j}) = 200\sqrt{2} \cdot 10^2 (\hat{i} + \hat{j}) = 200\sqrt{2} (\hat{i} + \hat{j}) \, N/C \]

Final Answer: \( 200\sqrt{2}(\hat{i} + \hat{j}) \, {N/C} \)