Question

Define electric field. Write its SI unit. A point charge \(q = 25\,\mu\text{C}\) is located at the point \((0, 0)\) in x-y plane. Find (i) the magnitude of electric field at point \((3\,\text{cm}, 4\,\text{cm})\), and (ii) the angle the field makes with the x-axis.

Hint:

Find \(r = 5\) cm from the coordinates. Use \(E = kq/r^2\). For a positive charge the field points along the line from the charge, so \(\tan\theta = 4/3\).

Answer 1

Step 1: Define electric field:
The electric field at a point is the force per unit positive test charge placed there: \(\vec{E} = \vec{F}/q_0\) (with \(q_0\) very small).
Its SI unit is newton per coulomb (\(\text{N C}^{-1}\)), which equals volt per metre (\(\text{V m}^{-1}\)).

Step 2: Find the distance:
The charge is at the origin and the point is \((3\ \text{cm}, 4\ \text{cm})\). So
\[ r = \sqrt{3^2 + 4^2} = 5\ \text{cm} = 5 \times 10^{-2}\ \text{m} \]

Step 3: Part (i): Magnitude of the field:
\[ E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} = \frac{9 \times 10^{9} \times 25 \times 10^{-6}}{(5 \times 10^{-2})^2} = \frac{2.25 \times 10^{5}}{2.5 \times 10^{-3}} = 9 \times 10^{7}\ \text{N C}^{-1} \]

Step 4: Part (ii): Direction of the field:
The charge is positive, so the field points away from it, along the line from the origin to \((3, 4)\).
\[ \tan\theta = \frac{4}{3} \quad \Rightarrow \quad \theta \approx 53^{\circ} \]

Final Answer:
(i) \(E = 9 \times 10^{7}\ \text{N C}^{-1}\). (ii) The field makes an angle of about \(53^{\circ}\) with the x-axis. \[ \boxed{E = 9 \times 10^{7}\ \text{N C}^{-1},\quad \theta \approx 53^{\circ}} \]