Question

Two point charges $q_A = 3\,\mu\text{C}$ and $q_B = -3\,\mu\text{C}$ are located $20\,\text{cm}$ apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude $1.5\times10^{-9}\,\text{C}$ is placed at this point, what is the force experienced by the test charge?

Hint:

Both fields at the midpoint point the same way (from +q to -q), so add them. Then use \(F = qE\); the negative test charge feels force opposite to \(\vec E\).

Answer 1

Given: \(q_A = +3\,\mu\text{C} = 3\times10^{-6}\,\text{C}\), \(q_B = -3\,\mu\text{C} = -3\times10^{-6}\,\text{C}\), separation \(AB = 20\,\text{cm}\). O is the midpoint, so each charge is \(r = 10\,\text{cm} = 0.10\,\text{m}\) from O.

Step 1: Field of a point charge. The magnitude of the field due to a point charge is

\[ E = \frac{k\,q}{r^{2}}, \qquad k = 9\times10^{9}\,\text{N m}^{2}\,\text{C}^{-2}. \]

Step 2: Field at O due to \(q_A\).

\[ E_A = \frac{(9\times10^{9})(3\times10^{-6})}{(0.10)^{2}} \]

\[ E_A = \frac{2.7\times10^{4}}{0.01} \]

\[ E_A = 2.7\times10^{6}\,\text{N C}^{-1}, \text{ directed from A toward B (away from }+q_A). \]

Step 3: Field at O due to \(q_B\). The magnitude is the same since \(|q_B| = |q_A|\) and the distance is the same:

\[ E_B = 2.7\times10^{6}\,\text{N C}^{-1}, \text{ directed from O toward B (toward }-q_B). \]

Step 4: Add the two fields. Both \(E_A\) and \(E_B\) point in the same direction, from A toward B, so they add:

\[ E = E_A + E_B = 2.7\times10^{6} + 2.7\times10^{6} \]

\[ E = 5.4\times10^{6}\,\text{N C}^{-1}, \text{ directed from A to B.} \]

Step 5: Force on the test charge (part b). A negative test charge \(q_t = 1.5\times10^{-9}\,\text{C}\) (in magnitude) is placed at O. The magnitude of the force is

\[ F = q_t\,E = (1.5\times10^{-9})(5.4\times10^{6}) \]

\[ F = 8.1\times10^{-3}\,\text{N}. \]

Because the charge is negative, the force is opposite to \(\vec E\), i.e. directed from B toward A.

\[\boxed{E = 5.4\times10^{6}\,\text{N C}^{-1}\ (A \to B), \quad F = 8.1\times10^{-3}\,\text{N}\ (B \to A)}\]

Answer 2

Expert method: short-dipole superposition with explicit vectors.

Step 1: Place the x-axis along AB with O at the origin. Then \(q_A = +q\) sits at \(x = -0.10\,\text{m}\) and \(q_B = -q\) sits at \(x = +0.10\,\text{m}\), where \(q = 3\times10^{-6}\,\text{C}\). The unit vector \(\hat\imath\) points from A to B.

Step 2: Field of \(q_A\) at O. A positive charge pushes outward, so its field at O points along \(+\hat\imath\):

\[ \vec E_A = \frac{kq}{r^{2}}\,\hat\imath = \frac{(9\times10^{9})(3\times10^{-6})}{(0.10)^{2}}\,\hat\imath = +2.7\times10^{6}\,\hat\imath\ \text{N C}^{-1}. \]

Step 3: Field of \(q_B\) at O. A negative charge pulls inward, so its field at O points toward B, i.e. along \(+\hat\imath\):

\[ \vec E_B = +2.7\times10^{6}\,\hat\imath\ \text{N C}^{-1}. \]

Step 4: Superpose:

\[ \vec E = \vec E_A + \vec E_B = 5.4\times10^{6}\,\hat\imath\ \text{N C}^{-1}. \]

Step 5 (consistency with the dipole formula): A and B form a dipole of moment \(p = q\cdot d = (3\times10^{-6})(0.20) = 6\times10^{-7}\,\text{C m}\). On the equatorial-versus-axial idea, O is on the axis, and the axial field of a dipole at distance \(x\) from its centre (with \(x \gg\) half-length) is \(E = 2kp/x^{3}\). Here O is not far compared with the half-length, so we instead sum the two point fields directly, which is exactly Steps 2 to 4. The direct sum gives \(5.4\times10^{6}\,\text{N C}^{-1}\) from A to B.

Step 6: Force on \(q = -1.5\times10^{-9}\,\text{C}\):

\[ \vec F = q\vec E = (-1.5\times10^{-9})(5.4\times10^{6}\,\hat\imath) = -8.1\times10^{-3}\,\hat\imath\ \text{N}, \]

i.e. magnitude \(8.1\times10^{-3}\,\text{N}\) directed from B to A.

\[\boxed{\vec E = 5.4\times10^{6}\,\hat\imath\ \text{N C}^{-1}, \quad F = 8.1\times10^{-3}\,\text{N along } -\hat\imath}\]