Exercises — Define electric field and electric potential

WHAT tool: field of a point charge, $E=kQ/r^2$. WHY this and not $V$: the question asks "how hard is the shove and which way" — that is force-per-charge, a vector, so it is the field, not the potential. $E=\frac{(8.99\times 10^9)(4\times 10^{-6})}{(0.2{)}^2}=\frac{3.596\times 10^4}{0.04}\approx 8.99\times 10^5\text{N/C}.$ Direction: $Q$ is positive, so the field points radially away from the charge. $\boxed{E\approx 9.0\times 10^5\text{N/C, pointing away.}}$

WHAT tool: $V=kQ/r$. WHY: "energy per coulomb to arrive here" is a single number — a scalar — so we use potential, and it falls as $1/r$ (one power of $r$), not $1/r^2$. $V=\frac{(8.99\times 10^9)(4\times 10^{-6})}{0.2}=\frac{3.596\times 10^4}{0.2}\approx 1.80\times 10^5\text{V}.$ Sanity link: $V=E\times r=8.99\times 10^5\times 0.2=1.80\times 10^5\text{V}$. ✔ (potential is field times distance for a point charge, because $1/r^2\times r=1/r$.)

WHAT tool: rearrange the definition $\vec{E}=\vec{F}/q_0$ into $\vec{F}=q_0\vec{E}$. WHY: the field was defined as force per charge; multiplying back by the charge recovers the force. $F=q_{0}E=(5\times 10^{-9})(250)=1.25\times 10^{-6}\text{N}=1.25\mu \text{N}.$ Charge is positive, so force is along $\vec{E}$ → east.

$F=q_{0}E=(-2\times 10^{-9})(600)=-1.2\times 10^{-6}\text{N}.$ WHAT the minus means: the negative sign says the force points opposite to $\vec{E}$ → south. Magnitude is $1.2\mu \text{N}$. WHY negative charges reverse: field is defined for a positive test charge; a negative charge feels the mirror-image push.

WHAT tool: invert $E=kQ/r^2$ to solve for $r$. WHY: we know $E$ and want $r$, so solve $r=\sqrt{kQ/E}$. $r=\sqrt{\frac{kQ}{E}}=\sqrt{\frac{(8.99\times 10^9)(9\times 10^{-6})}{1.0\times 10^4}}=\sqrt{\frac{8.091\times 10^4}{1.0\times 10^4}}=\sqrt{8.091}\approx 2.84\text{m}.$

WHAT tool: $W_{field}=q(V_i-V_f)$. WHY: energy is $U=qV$; the work the field does equals the drop in potential energy, so $W=U_i-U_f=q(V_i-V_f)$. $W_{field}=(3\times 10^{-9})(20-5)=(3\times 10^{-9})(15)=4.5\times 10^{-8}\text{J}=45\text{nJ}.$ Positive charge going to lower potential → field does positive work (it helps the motion). This is exactly how a battery drives current.

WHAT: each charge is $r=0.2\text{m}$ from the midpoint. Compute each field magnitude, then add as vectors. $E_1=E_2=\frac{(8.99\times 10^9)(5\times 10^{-6})}{(0.2{)}^2}=\frac{4.495\times 10^4}{0.04}\approx 1.124\times 10^6\text{N/C}.$ WHAT IT LOOKS LIKE (red arrows in figure): $E_1$ points away from $Q_1$ → to the right; $E_2$ points away from $Q_2$ → to the left. They are equal and opposite, so they cancel. $\boxed{E_{net}=0\text{at the midpoint.}}$ WHY: field is a vector; equal pushes in opposite directions sum to zero.

WHAT: potential is a scalar — no direction, just add the numbers. $V=\frac{k{Q}_1}{r}+\frac{k{Q}_2}{r}=2\times \frac{(8.99\times 10^9)(5\times 10^{-6})}{0.2}=2\times 2.2475\times 10^5\approx 4.50\times 10^5\text{V}.$ The punchline: here $E=0$ but $V\ne 0$. Field and potential are genuinely different animals — one cancelled, the other piled up.

Field (vectors): $E_1$ points away from $+Q_1$ → to the right. $E_2$ points toward $-Q_2$ → also to the right. Same direction, they add: $E_{net}=2\times 1.124\times 10^6\approx 2.25\times 10^6\text{N/C, toward the negative charge.}$ Potential (scalars): $V=\frac{k(+5\times 10^{-6})}{0.2}+\frac{k(-5\times 10^{-6})}{0.2}=+2.2475\times 10^5-2.2475\times 10^5=0\text{V}.$ The mirror of L3.1/3.2: here $E\ne 0$ but $V=0$. Both "surprising" cases from the parent note appear back-to-back.

Potentials: $V_A=\frac{(8.99\times 10^9)(6\times 10^{-6})}{0.10}=5.394\times 10^5\text{V},V_B=\frac{(8.99\times 10^9)(6\times 10^{-6})}{0.30}=1.798\times 10^5\text{V}.$ Difference: $V_A-V_B=5.394\times 10^5-1.798\times 10^5=3.596\times 10^5\text{V}.$ Work by field: $W=q(V_A-V_B)=(2\times 10^{-9})(3.596\times 10^5)=7.19\times 10^{-4}\text{J}.$ WHY positive: the positive charge falls from high potential ($A$, close in) to low ($B$, far out), so the field pushes it along and does positive work.

WHAT chain: potential energy lost → kinetic energy gained. $q\Delta V=\frac{1}{2}m{v}^2$. WHY this tool: energy is conserved; the electric field's work becomes motion. $\frac{1}{2}m{v}^2=q\Delta V=(1.6\times 10^{-19})(100)=1.6\times 10^{-17}\text{J}.$ $v=\sqrt{\frac{2q\Delta V}{m}}=\sqrt{\frac{2(1.6\times 10^{-17})}{1.67\times 10^{-27}}}=\sqrt{1.916\times 10^{10}}\approx 1.38\times 10^5\text{m/s}.$

WHAT: between two positive charges the fields point in opposite directions, so somewhere they cancel. Let the null point be distance $x$ from $Q_1$; then it is $(0.30-x)$ from $Q_2$. Set magnitudes equal: $\frac{k{Q}_1}{x^2}=\frac{k{Q}_2}{(0.30-x{)}^2}\Rightarrow \frac{4}{x^2}=\frac{1}{(0.30-x{)}^2}.$ WHY square-root both sides: it collapses the quadratic into a clean linear equation. $\frac{2}{x}=\frac{1}{0.30-x}\Rightarrow 2(0.30-x)=x\Rightarrow 0.60=3x\Rightarrow x=0.20\text{m}.$ CHECK the geometry (figure): the null sits closer to the smaller charge $Q_2$ — correct, because the stronger charge $Q_1$ pushes the balance point away from itself. The null is at $x=0.20\text{m}$ from $Q_1$ (i.e. $0.10\text{m}$ from $Q_2$). Note: the other algebraic root ($2/x=-1/(0.30-x)$) gives $x=0.60\text{m}$, which is outside the segment where both fields point the same way — so it is not a cancellation point. Always keep the physically valid root.

WHAT: potential is a scalar sum; set it to zero. Let the point be $x$ from $Q_1$: $\frac{k{Q}_1}{x}+\frac{k{Q}_2}{0.30-x}=0\Rightarrow \frac{4}{x}=\frac{1}{0.30-x}.$ (The $-1$ moved to the other side and flipped sign.) WHY: we want the positive contribution to exactly balance the negative one. $4(0.30-x)=x\Rightarrow 1.20=5x\Rightarrow x=0.24\text{m}.$ Contrast with L5.1: field-zero was at $0.20\text{m}$, potential-zero at $0.24\text{m}$ — different points, because one balances vectors and the other balances numbers. That single contrast is the whole lesson of this note.