Question

Two point charges \(q_1 = 2.5 \times 10^{-7}\,\text{C}\) and \(q_2 = -2.5 \times 10^{-7}\,\text{C}\) are located at points \((0, 0, -15\,\text{cm})\) and \((0, 0, 15\,\text{cm})\) respectively. Find:

(a) the electric dipole moment of the system, and

(b) the magnitude and direction of the electric field at the origin \((0, 0, 0)\).

Hint:

The dipole moment is q times 2a, pointing from the negative to the positive charge. At the midpoint both fields point the same way, so they add.

Answer 1

Step 1: Data:
\(q_1=+2.5\times10^{-7}\) C is at \(z=-15\) cm and \(q_2=-2.5\times10^{-7}\) C is at \(z=+15\) cm. So the two charges form a dipole of charge magnitude \(q=2.5\times10^{-7}\) C, with half-length \(a=0.15\) m.

Step 2: Dipole moment (part a):
The dipole moment is \(p=q\times2a\), and its direction is from the negative charge to the positive charge.
\[ p=2.5\times10^{-7}\times0.30=7.5\times10^{-8}\ \text{C m} \]
The negative charge is at \(+z\) and the positive charge is at \(-z\). So \(\vec p\) points along the \(-z\) direction.

Step 3: Field at the origin (part b):
The origin is midway between the charges, at \(r=0.15\) m from each. We use \(E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}\) with \(\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\) N m\(^2\) C\(^{-2}\).
\[ E_1=E_2=\frac{9\times10^{9}\times2.5\times10^{-7}}{(0.15)^2}=\frac{2250}{0.0225}=1\times10^{5}\ \text{N C}^{-1} \]

Step 4: Direction of each field:
The positive charge \(q_1\) is at \(-z\). Its field at the origin points away from it, so along \(+z\).
The negative charge \(q_2\) is at \(+z\). Its field at the origin points towards it, so also along \(+z\).
Both fields point the same way, so they add.
\[ E=E_1+E_2=2\times10^{5}\ \text{N C}^{-1}\ \text{along }+z \]
This is opposite to the direction of \(\vec p\), as expected for a point between the charges of a dipole.

Final Answer:
\[ \boxed{p=7.5\times10^{-8}\ \text{C m along }-z,\quad E=2\times10^{5}\ \text{N C}^{-1}\ \text{along }+z} \]