Question

Four point charges $q_A = 2\,\mu\text{C}$, $q_B = -5\,\mu\text{C}$, $q_C = 2\,\mu\text{C}$, $q_D = -5\,\mu\text{C}$ are located at the corners of a square ABCD of side $10\,\text{cm}$. What is the force on a charge of $1\,\mu\text{C}$ placed at the centre of the square?

Hint:

The centre is equidistant from all four corners. Diagonally opposite charges are equal and lie on opposite sides of O, so their forces cancel in pairs. Net force is zero.

Answer 1

Four charges sit at the corners of a square and we want the force on a \(1\ \mu\text{C}\) charge at the centre.

Step 1: Set up the geometry. Label the square ABCD with side \(a=10\ \text{cm}\). The centre O is equidistant from all four corners. Each corner is at a distance equal to half the diagonal,
\[r=\frac{a\sqrt2}{2}=\frac{a}{\sqrt2}.\]
Importantly, the diagonals AC and BD are straight lines passing through O, so A and C are on opposite sides of O along one diagonal, and B and D are on opposite sides of O along the other diagonal.

Step 2: Pair up the diagonally opposite charges.
\(q_A=2\ \mu\text{C}\) at A and \(q_C=2\ \mu\text{C}\) at C are equal in magnitude and sign, and both lie at the same distance \(r\) from O but in exactly opposite directions.

Step 3: Force from the A-C pair on the central charge \(q_0=1\ \mu\text{C}\).
\[F_A=\frac{k\,q_A q_0}{r^2}\ \text{(directed along OA)},\qquad F_C=\frac{k\,q_C q_0}{r^2}\ \text{(directed along OC)}\]
Since \(q_A=q_C\) and the distances are equal, \(F_A=F_C\), but OA and OC point in opposite directions. They are equal and opposite, so they cancel:
\[\vec F_A+\vec F_C=0.\]

Step 4: Force from the B-D pair. Likewise \(q_B=q_D=-5\ \mu\text{C}\) lie at equal distance \(r\) on opposite sides of O along the other diagonal.
\[\vec F_B+\vec F_D=0\]
by the same equal-and-opposite cancellation.

Step 5: Add all contributions.
\[\vec F_{\text{net}}=(\vec F_A+\vec F_C)+(\vec F_B+\vec F_D)=0+0=0.\]

Step 6: Conclusion. By the symmetry of the square, the forces from each diagonal pair cancel exactly, so the net electrostatic force on the central charge is zero.

\[\boxed{F_{\text{net}}=0}\]

Answer 2

Expert approach: vector superposition with explicit position vectors.

Step 1: Put the origin at the centre O. Place the corners symmetrically so each is at distance \(r=a/\sqrt2\) from O. Choose unit vectors along the diagonals: let A be at \(+r\,\hat u\) and C at \(-r\,\hat u\) (same diagonal), and B at \(+r\,\hat v\) and D at \(-r\,\hat v\) (other diagonal), with \(\hat u\perp\hat v\).

Step 2: The force the central charge feels from a corner charge \(q_i\) is, by Coulomb's law in vector form,
\[\vec F_i=\frac{k\,q_i q_0}{r^2}\,\hat r_{i\to O}\]
where \(\hat r_{i\to O}\) is the unit vector pointing from the corner toward O (the sign of \(q_i q_0\) decides whether it is toward or away).

Step 3: Write the diagonal pairs. For A and C, the unit vectors from corner to centre are \(-\hat u\) and \(+\hat u\) respectively, and \(q_A=q_C\). Their sum is
\[\vec F_A+\vec F_C=\frac{k q_A q_0}{r^2}(-\hat u)+\frac{k q_C q_0}{r^2}(+\hat u)=0.\]

Step 4: For B and D the same algebra with \(\hat v\) and \(q_B=q_D\) gives
\[\vec F_B+\vec F_D=\frac{k q_B q_0}{r^2}\big((-\hat v)+(+\hat v)\big)=0.\]

Step 5: Superpose:
\[\vec F_{\text{net}}=\sum_i\vec F_i=0.\]
The result is independent of the actual values of the charges and of the sign convention; it follows purely from the central symmetry (each charge has an identical partner diametrically opposite). Hence the net force vanishes.

\[\boxed{\vec F_{\text{net}}=0}\]