Electric Field

What this figure shows. A positive point charge +Q at the centre, with an imaginary spherical surface of radius r drawn around it and several arrows, each labelled E, drawn radiating straight outward from +Q to points on that sphere. The figure shows that the electric field vector E has the SAME magnitude at every point on the sphere (since every point on it is the same distance r from +Q) but points radially OUTWARD, away from the centre, at each location -- visually establishing $E=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$ directed along the outward radius for a positive source charge (and inw …

What this figure shows. A graph with force F (or, equivalently, electric field E) plotted on the vertical axis against separation/distance r on the horizontal axis. The plotted curve is a smooth, steeply DECREASING inverse-square curve: very large near r close to zero, falling rapidly as r increases, and flattening out to approach (but never actually reach) zero as r becomes very large -- the standard qualitative $1/r^2$ shape, visually representing that both $F\propto1/r^2$ and $E\propto1/r^2$ share the identical functional form since $E=F/q_0$ differs from F only by the constant test charge $q_0$ …

What this figure shows. Panel (a) shows two large, flat, parallel charged plates (one positive, one negative) facing each other, with a set of straight, parallel, EQUALLY-SPACED field lines drawn running from the positive plate directly across to the negative plate in the region between them -- illustrating a UNIFORM field, where E has the same magnitude AND the same direction at every point in that region. Panel (b) shows a single point charge with field lines radiating outward from it in all directions like spokes, the lines growing progressively farther apart (less dense) with increasing distance from the charge -- illustrating a NON-uniform field, where E's magnitude is constant only on any one sphere of fixed radius r centred on the charge, while its DIRECTION differs from point to point (alw …

Worked out. Charges of $+10\,\mu C$ at A and $+10\,\mu C$ at C sit at the two ends of the hypotenuse of a right isosceles triangle with the right angle at B (legs $AB=BC=5$ cm), where $+2\,\mu C$ sits at B; P is the midpoint of hypotenuse AC. Since P is equidistant from A and C ($AP=CP=\frac{AC}{2}\approx3.54$ cm, using $AC=5\sqrt2$ cm) and A, C carry EQUAL charges, the fields $\vec{E}_A$ and $\vec{E}_C$ at P are equal in magnitude and exactly opposite in direction, so they CANCEL completely, leaving only the field due to the $2\,\mu C$ charge at B. Using the right-triangle property that the median from the right angle to the hypotenuse equals half the hypotenuse, $BP=AP=CP\approx3.54$ cm, so $E_B=\frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{(0.0354)^2}\approx1.44\times10^7$ N/C, directed from B straight through P (along BP) -- the example is a clean illustration of using symmetry to eliminate two of three fields before apply …

Worked out. In a simplified hydrogen-atom model, an electron orbits a proton (charge $+1.6\times10^{-19}$ C) at distance $r=5.3\times10^{-11}$ m. The electric field due to the proton at this distance is $E=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}=9\times10^9\times\frac{1.6\times10^{-19}}{(5.3\times10^{-11})^2}\approx5.1\times10^{11}$ N/C. The force on the orbiting electron (charge $-1.6\times10^{-19}$ C) is then $F=qE=(-1.6\times10^{-19})(5.1\times10^{11})\approx-8.16\times10^{-8}$ N, i.e. attractive, pulling the electron toward the proton; the same result is cross-checked directly via Coulomb's law, $F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\approx8.2\times10^{-8}$ N, confirming that finding the field first and then multiplying by a second charge (F=qE) gives exactly the same force as applying Coulomb's law to the …