Net electric field at the center of a charged octagon

Key concept: This problem is solved by symmetry and vector addition of fields from opposite vertices.

Step by step

Use symmetry first

At the center of a regular octagon, each charge is the same distance $R$ from the center. The electric field from each charge has magnitude

$E_0=\frac{k|Q|}{R^2}.$

The direction is along the line joining the vertex and the center: away from a positive charge and toward a negative charge.

Because the four upper charges are $+Q$ and the four lower charges are $-Q$, the horizontal components cancel by left-right symmetry. What remains is a vertical net field.

Direction of the field

A positive charge in the upper half pushes the field at the center downward. A negative charge in the lower half pulls the field at the center downward as well. So every charge contributes a downward vertical component.

For a regular octagon, the eight field vectors are arranged symmetrically. The horizontal components cancel in pairs, and the vertical components from all eight charges add in the downward direction. The exact sum gives a net field straight downward.

Magnitude in terms of $R$

If the octagon is oriented so that one pair of opposite vertices is horizontal and the upper four vertices carry $+Q$, then the vertical component sum is

$E_{\text{net}}=\sqrt{2}\,\frac{kQ}{R^2}$

directed downward.

This can be checked by resolving each vertex field into components using the octagon angles: the contributions from the top and bottom pairs combine so that the net vertical coefficient is $\sqrt{2}$.

Rewrite in terms of side length $L$

For a regular octagon,

$R=\frac{L}{2\sin(\pi/8)}.$

Substitute this into the result:

=\sqrt{2}\,kQ\,\frac{4\sin^2(\pi/8)}{L^2}.$$ Using $\sin^2(\pi/8)=\frac{2-\sqrt2}{4}$, this simplifies to $$E_{\text{net}}=\frac{kQ(2-\sqrt2)}{L^2}\,\sqrt{2}.$$ ## If all eight charges had the same sign If all eight charges were the same sign, the field at the center would be zero by symmetry. Opposite vertices would produce equal and opposite field vectors, so the vector sum cancels completely. ## Final answer - **Magnitude:** $\boxed{\sqrt{2}\,\dfrac{kQ}{R^2}}$ - **Direction:** downward - **In terms of $L$:** substitute $R=\dfrac{L}{2\sin(\pi/8)}$ - **All same sign:** the net field at $O$ is zero ### Pitfall alert A common mistake is to add the field magnitudes as if they were scalar quantities. Electric field is a vector, so direction matters more than raw size. Another error is to forget that a negative charge reverses the field direction at the center. In this problem, the lower negative charges do not create upward field at the center; they still contribute downward because the field points toward a negative charge. Symmetry should be used before doing any lengthy trig work. ### Try different conditions If the octagon were rearranged so that the alternating charges were placed as $+Q,-Q,+Q,-Q,\ldots$ around the vertices, the symmetry would change and the net field might point along a different axis or cancel entirely depending on the orientation. If the charges were all $+Q$, then opposite vertices would always cancel pairwise at the center, giving zero net field without any trigonometric calculation. The key variation is the charge pattern, not the polygon shape alone. ### Further reading electric field vector, regular polygon symmetry, superposition principle