Electric Field on a Corner Electron of a Charged Square: 04-BS-9, May 2015, Q1

Question 1: Electric Field on a Corner Electron of a Charged Square (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Square side $a$	$10^{-10}$ m
Corner charges	four electrons, $-e$ each, one per corner
Centre charge	four protons collocated at the centre, total $+4e$

Find. The magnitude and direction of $\vec E$ at the location of one electron.

Four electrons at the corners of a 10^-10 m square; four protons (net +4e) collocated at the centre. Field is evaluated at one corner electron.

Approach. Superpose the Coulomb field from the +4e centre charge and from the other three corner electrons at the location of the target electron. By symmetry, the two "adjacent" electrons' off-diagonal components cancel identically, leaving a resultant purely along the corner's own diagonal.

Distance from centre to a corner (half-diagonal). $$r_c=\frac{a\sqrt2}{2}=\frac{(10^{-10})\sqrt2}{2}$$ $$r_c=\boxed{7.071\times10^{-11}\ \text{m}}$$
Field from the +4e centre charge, at the target corner. Directed radially outward from the centre, along the diagonal: $$E_{\text{centre}}=\frac{k(4e)}{r_c^2}=\frac{(8.988\times10^9)(4\times1.6\times10^{-19})}{(7.071\times10^{-11})^2}$$ $$E_{\text{centre}}=\boxed{1.1511\times10^{12}\ \text{V/m, outward}}$$
Superpose the other three electrons (each $-e$, field pointing toward each source) and resolve into components. The diagonally-opposite electron (distance $r_c$) pulls the field back inward along the same diagonal; the two adjacent electrons (distance $a$) each contribute a field toward themselves, which by symmetry combine to leave only a diagonal component. Carrying out the full vector sum (both Cartesian components equal, i.e. exactly $45^\circ$): $$E_x=E_y=6.191\times10^{11}\ \text{V/m}$$ $$E=\sqrt{E_x^2+E_y^2}=\boxed{8.756\times10^{11}\ \text{V/m}}$$ directed at $45^\circ$ — i.e. radially outward from the centre, along the diagonal through that electron.

Quantity	Result
$r_c$ (centre to corner)	$7.071\times10^{-11}$ m
$E$ at the corner electron	$8.756\times10^{11}$ V/m, outward along the diagonal

Concept & Theory

Topic. Superposition of Coulomb fields from a discrete point-charge cluster.

Key relations. $\vec E=\sum_i \dfrac{kq_i}{r_i^2}\hat r_i$, summed as vectors, not magnitudes.

Why this works. Coulomb's law is linear in each source charge, so the net field at any point is the vector sum of the fields each charge would produce alone. The dominant +4e centre charge, being both larger and closer than any of the three remaining $-e$ corners, sets the overall outward sense of the resultant, but the other electrons' fields still shift its magnitude and must be summed as vectors, not simply subtracted.

Common pitfall. Adding the four contributions as scalars (magnitudes) instead of vectors, or forgetting that the four protons sit at one collocated point (net $+4e$), not spread individually.

Source. Sadiku, Elements of Electromagnetics, Ch. 4 (Coulomb's Law and superposition of point charges).

Question 1 of 8: Electric Field on a Corner Electron of a Charged Square

Question 1: Electric Field on a Corner Electron of a Charged Square (20 marks)