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Derivation

Electric Field from Coulomb's Law

D-039 Home PU-102 Threads force · fields Depends on superposition-of-forces
Statement

The electrostatic field \(\vec{E}(\vec{r})\) is defined as the electric force per unit stationary test charge in the limit that the test charge is vanishingly small, \(\vec{E}(\vec{r}) = \lim_{q\to 0}\vec{F}(\vec{r})/q\). Combining this definition with Coulomb's law and the superposition of forces yields, for a source charge density \(\rho(\vec{r}')\) at rest, \(\displaystyle \vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\int \rho(\vec{r}')\,\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^{3}}\,d^{3}r'\).

Why it matters

The field concept removes the test charge from the problem entirely: instead of asking "what force does this configuration exert on that charge," we assign to every point of space a vector \(\vec{E}(\vec{r})\) that depends only on the source charges. Any charge \(q\) placed at \(\vec{r}\) then feels \(\vec{F}=q\vec{E}(\vec{r})\). This is the conceptual pivot from action-at-a-distance to local field theory.

The superposition integral is the master formula of electrostatics. Every later result — Gauss's law, the potential \(V\), boundary-value problems, multipole expansions — is either derived from it or is a computational shortcut for evaluating it. Getting its structure exactly right (the inverse-square magnitude carried by an inverse-cube vector, the source point \(\vec{r}'\) versus the field point \(\vec{r}\)) is foundational.

Assumptions
Sources are static.If the charges move, magnetic fields appear and the force acquires a velocity-dependent term \(q\vec{v}\times\vec{B}\); the pure \(1/r^2\) Coulomb form and the time-independent field both fail, and one needs the full Maxwell (Liénard–Wiechert) treatment.
The test charge is infinitesimal.A finite test charge polarizes or displaces the sources (image effects, redistribution on conductors), so the measured \(\vec{F}/q\) would depend on \(q\) and would not equal the source-only field. The \(\lim_{q\to0}\) makes \(\vec{E}\) a property of the sources alone.
Superposition of forces holds.Without linear additivity of Coulomb forces the integral over the distribution is meaningless; the field of the whole would not be the sum of the fields of the parts. This is an empirical linearity of Maxwell's equations in vacuum.
The charge density \(\rho(\vec{r}')\) is integrable and the field point is treated with care inside the source.If \(\vec{r}\) lies within a continuous distribution the integrand has a \(1/|\vec{r}-\vec{r}'|^2\) singularity; the integral still converges (the volume element supplies \(r'^2\,dr'\)), but naive point-charge reasoning at \(\vec{r}=\vec{r}'\) diverges. For genuine point charges \(\vec{E}\) is undefined at the charge itself.
We work in vacuum (or absorb the medium into \(\varepsilon_0\to\varepsilon\)).In a polarizable medium the bound charges contribute; one must distinguish \(\vec{E}\) from \(\vec{D}\), and \(\varepsilon_0\) is replaced by the permittivity \(\varepsilon\) only for linear, homogeneous, isotropic media.
Derivation
1
\[ \vec{F}_{1}(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\,\frac{q\,q_1}{|\vec{r}-\vec{r}_1|^{2}}\,\hat{\mathcal{R}}_1, \qquad \hat{\mathcal{R}}_1 \equiv \frac{\vec{r}-\vec{r}_1}{|\vec{r}-\vec{r}_1|}. \]
Coulomb's law for the force on a test charge \(q\) at \(\vec{r}\) due to a single point source \(q_1\) at \(\vec{r}_1\); the unit vector points from source to test charge (repulsion for like signs). A
2
\[ \vec{F}(\vec{r}) = \sum_{i=1}^{N}\frac{1}{4\pi\varepsilon_0}\,\frac{q\,q_i}{|\vec{r}-\vec{r}_i|^{2}}\,\frac{\vec{r}-\vec{r}_i}{|\vec{r}-\vec{r}_i|}. \]
Apply the assumed superposition of forces: the net force from \(N\) sources is the vector sum of the individual Coulomb forces. Combining the \(1/|\vec{r}-\vec{r}_i|^2\) magnitude with the unit vector produces the inverse-cube form. A
3
\[ \vec{E}(\vec{r}) \equiv \lim_{q\to 0}\frac{\vec{F}(\vec{r})}{q} = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{N} q_i\,\frac{\vec{r}-\vec{r}_i}{|\vec{r}-\vec{r}_i|^{3}}. \]
Divide by the test charge and take the limit \(q\to0\) so the sources are not disturbed. The factor \(q\) cancels linearly, leaving a quantity that depends only on the source charges and geometry — this defines the field. A
4
\[ q_i \;\longrightarrow\; dq' = \rho(\vec{r}')\,d^{3}r', \qquad \sum_i \;\longrightarrow\; \int_{V} . \]
Pass to a continuous distribution: partition the source region into cells of volume \(d^3r'\) carrying charge \(dq'=\rho\,d^3r'\). Because superposition is linear, the Riemann sum of point-charge fields converges to an integral as the cells shrink. B
5
\[ \vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\int_{V} \rho(\vec{r}')\,\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^{3}}\,d^{3}r'. \]
Substitute the continuum replacement into step 3. The integration runs over the primed (source) coordinates while \(\vec{r}\) (the field point) is held fixed — the separation vector \(\vec{r}-\vec{r}'\) is the "from source to field point" direction inherited from step 1. B
6
\[ \vec{E}(\vec{r}) = -\frac{1}{4\pi\varepsilon_0}\int_{V}\rho(\vec{r}')\,\nabla\!\left(\frac{1}{|\vec{r}-\vec{r}'|}\right)d^{3}r' = -\nabla\Phi(\vec{r}),\quad \Phi=\frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}\,d^{3}r'. \]
Optional but important: since \(\dfrac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^{3}} = -\nabla_{\vec r}\dfrac{1}{|\vec{r}-\vec{r}'|}\) and \(\nabla\) acts on the unprimed variable (so it passes through the \(d^3r'\) integral), the electrostatic field is the gradient of a scalar potential. This proves \(\nabla\times\vec{E}=0\) for any static distribution. C
Result
\[ \boxed{\;\vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\int \rho(\vec{r}')\,\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^{3}}\,d^{3}r'\;}\qquad \vec{F} = q\,\vec{E}(\vec{r}) \]

Reading. Every element of source charge \(dq'=\rho\,d^3r'\) contributes a little Coulomb field pointing radially away from itself (for \(dq'>0\)), falling off as the inverse square of the distance; the total field is the vector superposition of all these contributions. The field is a property of the sources alone — the test charge has been divided out. A charge \(q\) dropped at \(\vec{r}\) then feels \(\vec{F}=q\vec{E}\).

Units check. \(\left[\dfrac{1}{4\pi\varepsilon_0}\right]=\dfrac{\text{N}\cdot\text{m}^2}{\text{C}^2}\), \([\rho\,d^3r']=\text{C}\), and \(\left[\dfrac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^{3}}\right]=\text{m}^{-2}\). Product: \(\dfrac{\text{N}\,\text{m}^2}{\text{C}^2}\cdot\text{C}\cdot\text{m}^{-2}=\dfrac{\text{N}}{\text{C}}=\dfrac{\text{V}}{\text{m}}\). Correct dimensions of field strength.

Limiting cases
  • Single point charge. \(\rho(\vec{r}')=Q\,\delta^{3}(\vec{r}'-\vec{r}_0)\) collapses the integral to \(\vec{E}=\dfrac{Q}{4\pi\varepsilon_0}\dfrac{\vec{r}-\vec{r}_0}{|\vec{r}-\vec{r}_0|^{3}}\), recovering Coulomb's law as a field.
  • Far field of a bounded neutral distribution. For \(r\gg\) source size with total charge zero, the monopole term vanishes and \(\vec{E}\sim 1/r^{3}\) (dipole) — the leading nonzero multipole dominates.
  • Far field of a net-charged distribution. For \(r\gg\) source size, \(\vec{E}\to \dfrac{Q_{\text{tot}}}{4\pi\varepsilon_0 r^{2}}\hat{r}\): from far away any lump of total charge \(Q_{\text{tot}}\) looks like a point charge.
  • Continuous limit / smooth interior. Inside a smooth \(\rho\), the field is finite and continuous; only surface charge layers produce a discontinuity \(\Delta E_\perp=\sigma/\varepsilon_0\).
Breaks when
  • Moving or accelerating sources. A time-dependent \(\rho(\vec{r}',t)\) radiates; the instantaneous Coulomb integral is wrong because information propagates at \(c\). One must use retarded time and include the magnetic field — the electrostatic formula is only the \(v/c\to0\), static limit.
  • Evaluation exactly at a point charge. At \(\vec{r}=\vec{r}_i\) for a genuine point source the integrand diverges and \(\vec{E}\) is undefined (formally infinite self-field). The field is only defined at points away from point charges; the classical self-energy is divergent.
  • Strong fields / quantum regime. Near \(\sim 10^{18}\,\text{V/m}\) (Schwinger scale) the vacuum becomes nonlinear (pair production, QED corrections) and classical superposition fails; the linear \(\vec{F}=q\vec{E}\) response of the "test charge" also breaks down.
  • Polarizable / conducting media without correction. If the region between sources is a dielectric or conductor, bound and induced charges are not in \(\rho\); using only the free charge gives the wrong field unless one solves the boundary-value problem or replaces \(\varepsilon_0\to\varepsilon\).
Failure modes
  • Confusing field point and source point. Integrating over \(\vec{r}\) instead of \(\vec{r}'\), or writing the separation as \(\vec{r}'-\vec{r}\) (wrong sign, field points toward positive charge). The primed variable is dummy; \(\vec{r}\) is fixed.
  • Using \(1/|\vec{r}-\vec{r}'|^{2}\) with a full vector \((\vec{r}-\vec{r}')\). That double-counts one power of distance. Either \(\dfrac{\hat{\mathcal R}}{|\cdots|^2}\) or \(\dfrac{\vec{r}-\vec{r}'}{|\cdots|^{3}}\), never \(\dfrac{\vec{r}-\vec{r}'}{|\cdots|^{2}}\).
  • Adding magnitudes instead of vectors. Superposition is a vector sum; summing \(|\vec{E}_i|\) overestimates the field wherever contributions partially cancel (e.g. on the axis of a dipole vs. between the charges).
  • Forgetting the \(q\to0\) limit / treating the test charge as physical. Plugging in a finite probe charge that itself alters the sources (especially near conductors) gives a self-consistent force, not the source field.
  • Dropping the vector nature of \(d\vec{E}\) by symmetry too early. Cancelling components "by inspection" without checking the symmetry axis, e.g. keeping a component off-axis where it does not vanish.
  • Wrong volume element for the geometry. Using \(dx'\) for a line, \(dA'\) for a surface, or forgetting the \(r'^2\sin\theta'\) Jacobian in spherical coordinates.
Discussion

The deepest content of this derivation is that the definition \(\vec{E}=\lim_{q\to0}\vec{F}/q\) turns force — a relation between two charges — into a field, a property of space sourced by one set of charges. This is not merely bookkeeping. Once we accept that \(\vec{E}\) exists at every point whether or not a test charge is there, we can ask about the field's own dynamics, its energy density \(u=\tfrac12\varepsilon_0 E^2\), and its momentum. The static Coulomb integral is the seed of a full field theory in which \(\vec{E}\) and \(\vec{B}\) are the true dynamical objects.

Structurally, the superposition integral says the map \(\rho\mapsto\vec{E}\) is linear. Linearity is what lets us build complicated fields from simple ones, expand in multipoles, and use Green's functions: the integrand \(\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3}\) is precisely the field of a unit point source, i.e. the Green's function of electrostatics (up to a gradient). Solving any electrostatics problem is, in principle, convolving this kernel with \(\rho\).

Step 6 exposed a hidden gift: because the Coulomb kernel is a gradient, \(\vec{E}=-\nabla\Phi\) and therefore \(\nabla\times\vec{E}=0\) for every static distribution. The curl-free property is not an extra assumption; it is a theorem that follows from the \(1/r^2\) law plus superposition. This is why electrostatic problems reduce to a single scalar potential and to Poisson's equation \(\nabla^2\Phi=-\rho/\varepsilon_0\), the differential counterpart of the integral derived here.

Taking the divergence of the boxed result and using \(\nabla\cdot\dfrac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3}=4\pi\,\delta^{3}(\vec{r}-\vec{r}')\) collapses the integral to \(\nabla\cdot\vec{E}=\rho/\varepsilon_0\) — Gauss's law in differential form. Thus Coulomb's law plus superposition is logically equivalent to Gauss's law (given the curl-free condition): the integral form derived here and the local Maxwell equation are two faces of the same physics, related by the delta-function identity that regularizes the point-source singularity. The subtlety is entirely in that identity: the field of a point charge is smooth everywhere except at the source, where its divergence is a delta function carrying exactly the enclosed charge.

Common misconceptions. (i) The field does not "know" about the test charge — reversing the sign of \(q\) reverses the force, not the field. (ii) \(\vec{E}\) points away from positive source charge and toward negative source charge, independent of the sign of any charge you later place there. (iii) A vanishing field at a point does not imply vanishing charge nearby (e.g. the midpoint between two equal like charges), and a vanishing potential does not imply a vanishing field.

Worked examples
1
Field on the axis of a uniformly charged ring. A ring of radius \(a\) carries total charge \(Q\) uniformly, so linear density \(\lambda=Q/2\pi a\). Find \(\vec{E}\) a distance \(z\) along the axis. A
\[ d\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{\lambda\,dl'}{(a^2+z^2)}\,\hat{\mathcal R},\qquad \hat{\mathcal R}\cdot\hat{z}=\frac{z}{\sqrt{a^2+z^2}}. \]
By symmetry the radial components from opposite arcs cancel; only the axial component survives. Multiply the magnitude by the axial projection.
\[ E_z=\frac{1}{4\pi\varepsilon_0}\frac{\lambda}{(a^2+z^2)}\frac{z}{\sqrt{a^2+z^2}}\oint dl' =\frac{1}{4\pi\varepsilon_0}\frac{\lambda(2\pi a)\,z}{(a^2+z^2)^{3/2}} =\frac{1}{4\pi\varepsilon_0}\frac{Q\,z}{(a^2+z^2)^{3/2}}. \]
The circumference integral \(\oint dl'=2\pi a\) gives \(\lambda\,2\pi a=Q\). Now insert numbers: \(Q=5.0\,\text{nC}=5.0\times10^{-9}\,\text{C}\), \(a=4.0\,\text{cm}=0.040\,\text{m}\), \(z=3.0\,\text{cm}=0.030\,\text{m}\), \(\tfrac{1}{4\pi\varepsilon_0}=8.99\times10^{9}\,\text{N·m}^2/\text{C}^2\).
\[ (a^2+z^2)^{3/2}=(0.0016+0.0009)^{3/2}=(0.0025)^{3/2}=1.25\times10^{-4}\,\text{m}^3. \]
\[ E_z=8.99\times10^{9}\cdot\frac{(5.0\times10^{-9})(0.030)}{1.25\times10^{-4}} =8.99\times10^{9}\cdot\frac{1.5\times10^{-10}}{1.25\times10^{-4}}. \]
\[ E_z \approx 1.08\times10^{4}\ \text{N/C}\quad(\text{along }+\hat z). \]

Reading. The field pushes a positive test charge away from the ring along the axis. Note \(E_z\to0\) both at \(z=0\) (symmetry) and as \(z\to\infty\) (\(\sim Q/4\pi\varepsilon_0 z^2\)); it peaks at \(z=a/\sqrt2\).

Units check. \(\dfrac{\text{N·m}^2}{\text{C}^2}\cdot\dfrac{\text{C·m}}{\text{m}^3}=\dfrac{\text{N}}{\text{C}}\). Correct.

2
Field above the center of a uniformly charged disk. A disk of radius \(R\) carries uniform surface density \(\sigma\). Find \(\vec{E}\) at height \(z\) on the axis by superposing rings. B
\[ dE_z=\frac{1}{4\pi\varepsilon_0}\frac{(\sigma\,2\pi a\,da)\,z}{(a^2+z^2)^{3/2}}, \]
Treat the disk as nested rings of radius \(a\), width \(da\), charge \(dQ=\sigma\,2\pi a\,da\); each contributes the ring result from Example 1. Integrate \(a:0\to R\).
\[ E_z=\frac{\sigma z}{2\varepsilon_0}\int_0^{R}\frac{a\,da}{(a^2+z^2)^{3/2}} =\frac{\sigma z}{2\varepsilon_0}\left[-\frac{1}{\sqrt{a^2+z^2}}\right]_0^{R} =\frac{\sigma}{2\varepsilon_0}\left(1-\frac{z}{\sqrt{R^2+z^2}}\right). \]
Substitute \(u=a^2+z^2\), \(du=2a\,da\); the antiderivative is \(-u^{-1/2}\). Numbers: \(\sigma=2.0\times10^{-6}\,\text{C/m}^2\), \(R=0.10\,\text{m}\), \(z=0.050\,\text{m}\), \(\varepsilon_0=8.85\times10^{-12}\,\text{C}^2/\text{N·m}^2\).
\[ \frac{\sigma}{2\varepsilon_0}=\frac{2.0\times10^{-6}}{2(8.85\times10^{-12})}=1.13\times10^{5}\ \text{N/C},\qquad \frac{z}{\sqrt{R^2+z^2}}=\frac{0.050}{\sqrt{0.0125}}=\frac{0.050}{0.1118}=0.447. \]
\[ E_z=1.13\times10^{5}(1-0.447)\approx 6.2\times10^{4}\ \text{N/C}. \]

Reading. As \(R\to\infty\) (or \(z\to0^+\)) the bracket \(\to1\) and \(E_z\to\sigma/2\varepsilon_0\), the field of an infinite sheet — independent of distance. Here the finite disk gives about 55% of that plateau value.

Units check. \(\dfrac{\text{C/m}^2}{\text{C}^2/\text{N·m}^2}=\dfrac{\text{N}}{\text{C}}\); the bracket is dimensionless. Correct.

Problems
  1. A point charge \(Q=+2.0\,\mu\text{C}\) sits at the origin. Find the magnitude and direction of \(\vec{E}\) at the point \((0.30,\,0.40,\,0)\,\text{m}\).
    Solution \(r=\sqrt{0.30^2+0.40^2}=0.50\,\text{m}\). \(E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}=8.99\times10^{9}\cdot\dfrac{2.0\times10^{-6}}{0.25}=7.19\times10^{4}\,\text{N/C}\). Direction \(\hat r=(0.6,0.8,0)\), pointing radially outward from the origin. So \(\vec{E}\approx(4.3,\,5.8,\,0)\times10^{4}\,\text{N/C}\).
  2. Two charges \(+q\) and \(-q\) (a dipole) sit at \(x=\pm d/2\) on the \(x\)-axis. Show that on the perpendicular bisector (the \(y\)-axis) at distance \(y\gg d\) the field is \(E\approx\dfrac{1}{4\pi\varepsilon_0}\dfrac{qd}{y^3}\), directed along \(-\hat x\), and evaluate for \(q=3.0\,\text{nC}\), \(d=2.0\,\text{mm}\), \(y=0.10\,\text{m}\).
    Solution Each charge is a distance \(s=\sqrt{y^2+(d/2)^2}\) away. The \(y\)-components cancel; the \(x\)-components add. Each contributes \(E_1=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{s^2}\) with \(x\)-projection \((d/2)/s\). Total \(E_x=2\cdot\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{s^2}\dfrac{d/2}{s}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{qd}{s^3}\). Both \(x\)-components point from \(+q\) toward \(-q\), i.e. \(-\hat x\). For \(y\gg d\), \(s\to y\): \(E\approx\dfrac{1}{4\pi\varepsilon_0}\dfrac{qd}{y^3}\). Number: \(p=qd=6.0\times10^{-12}\,\text{C·m}\); \(E=8.99\times10^{9}\cdot\dfrac{6.0\times10^{-12}}{(0.10)^3}=8.99\times10^{9}\cdot6.0\times10^{-9}=54\,\text{N/C}\), along \(-\hat x\).
  3. An infinite line charge has uniform density \(\lambda\). Using the superposition integral, show \(E(s)=\dfrac{\lambda}{2\pi\varepsilon_0 s}\) at perpendicular distance \(s\). Evaluate for \(\lambda=4.0\times10^{-8}\,\text{C/m}\), \(s=0.020\,\text{m}\).
    Solution Put the wire on the \(z\)-axis, field point at distance \(s\). Element \(dz'\) at height \(z'\) is a distance \(\sqrt{s^2+z'^2}\) away; by symmetry only the radial component survives, with projection \(s/\sqrt{s^2+z'^2}\). \(E_s=\dfrac{\lambda}{4\pi\varepsilon_0}\displaystyle\int_{-\infty}^{\infty}\dfrac{s\,dz'}{(s^2+z'^2)^{3/2}}\). The integral equals \(\left[\dfrac{z'}{s^2\sqrt{s^2+z'^2}}\right]_{-\infty}^{\infty}=\dfrac{2}{s^2}\). Hence \(E_s=\dfrac{\lambda}{4\pi\varepsilon_0}\cdot\dfrac{s\cdot 2}{s^2}=\dfrac{\lambda}{2\pi\varepsilon_0 s}\). Number: \(E=\dfrac{4.0\times10^{-8}}{2\pi(8.85\times10^{-12})(0.020)}=\dfrac{4.0\times10^{-8}}{1.112\times10^{-12}}\approx3.6\times10^{4}\,\text{N/C}\).
  4. Return to the ring of Example 1 (\(Q=5.0\,\text{nC}\), \(a=4.0\,\text{cm}\)). At what axial distance \(z\) is \(E_z\) maximum, and what is that maximum value?
    Solution Set \(\dfrac{dE_z}{dz}=0\) with \(E_z\propto z(a^2+z^2)^{-3/2}\): \(\dfrac{d}{dz}\big[z(a^2+z^2)^{-3/2}\big]=(a^2+z^2)^{-3/2}-3z^2(a^2+z^2)^{-5/2}=0\Rightarrow a^2+z^2=3z^2\Rightarrow z=a/\sqrt2\). So \(z=0.040/\sqrt2=0.0283\,\text{m}\). Then \((a^2+z^2)^{3/2}=(1.5a^2)^{3/2}=1.5^{3/2}a^3=1.837\,(0.040)^3=1.176\times10^{-4}\,\text{m}^3\). \(E_{z,\max}=8.99\times10^{9}\cdot\dfrac{(5.0\times10^{-9})(0.0283)}{1.176\times10^{-4}}\approx1.08\times10^{4}\,\text{N/C}\). (Nearly equal to the \(z=3\,\text{cm}\) value, since \(3\,\text{cm}\) is close to the maximum at \(2.83\,\text{cm}\).)
  5. A solid sphere of radius \(R\) carries uniform volume charge density \(\rho\). Using the equivalence of the superposition integral with Gauss's law, find \(\vec{E}\) both inside (\(r<R\)) and outside (\(r>R\)), and evaluate the surface field for \(\rho=1.0\times10^{-6}\,\text{C/m}^3\), \(R=0.050\,\text{m}\).
    Solution By spherical symmetry \(\vec{E}=E(r)\hat r\). A Gaussian sphere of radius \(r\) gives \(E\,4\pi r^2=Q_{\text{enc}}/\varepsilon_0\). Outside: \(Q_{\text{enc}}=\tfrac43\pi R^3\rho\), so \(E=\dfrac{\rho R^3}{3\varepsilon_0 r^2}\) (\(=\dfrac{Q_{\text{tot}}}{4\pi\varepsilon_0 r^2}\), like a point charge). Inside: \(Q_{\text{enc}}=\tfrac43\pi r^3\rho\), so \(E=\dfrac{\rho r}{3\varepsilon_0}\), growing linearly from the center. At the surface \(r=R\): \(E=\dfrac{\rho R}{3\varepsilon_0}=\dfrac{(1.0\times10^{-6})(0.050)}{3(8.85\times10^{-12})}=\dfrac{5.0\times10^{-8}}{2.655\times10^{-11}}\approx1.9\times10^{3}\,\text{N/C}\), directed radially outward.