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The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Chapter 23 — Electric Potential

Problems with worked solutions · 12 points · 3 extra credit

Chapter 23 — Electric Potential

Core: 12 points · Extra credit: 3 points. Show your reasoning; partial credit is given. Use k=8.99×109 N⋅m2/C2k=8.99\times10^{9}\,\text{N}\cdot\text{m}^2/\text{C}^2, e=1.60×10−19 Ce=1.60\times10^{-19}\,\text{C}, me=9.11×10−31 kgm_e=9.11\times10^{-31}\,\text{kg}, mp=1.67×10−27 kgm_p=1.67\times10^{-27}\,\text{kg}.

  1. 1. (2 pts ••) A charge q1=+4.0 nCq_1=+4.0\,\text{nC} is at the origin and q2=−3.0 nCq_2=-3.0\,\text{nC} is at x=0.50 mx=0.50\,\text{m}.

    (a) Find the electric potential at the midpoint between them (x=0.25 mx=0.25\,\text{m}), taking V=0V=0 at infinity.

    (b) Find the work required to bring a +2.0 nC+2.0\,\text{nC} charge from infinity to that midpoint.

    Solution

    1.

    (a) Both charges are 0.25 m0.25\,\text{m} from the midpoint: V=k(q1+q2)r=(8.99×109)(1.0×10−9)0.25≈36 VV=\dfrac{k(q_1+q_2)}{r}=\dfrac{(8.99\times10^{9})(1.0\times10^{-9})}{0.25}\approx 36\,\text{V}.

    (b) W=q ΔV=qVP=(2.0×10−9)(36)≈7.2×10−8 JW=q\,\Delta V=qV_P=(2.0\times10^{-9})(36)\approx 7.2\times10^{-8}\,\text{J} (with V=0V=0 at infinity).

  2. 2. (2 pts ••) A proton is released from rest at a distance of 2.0 cm2.0\,\text{cm} from a fixed point charge Q=+5.0 nCQ=+5.0\,\text{nC}. Find its speed when it has moved to 10.0 cm10.0\,\text{cm} from the charge.

    Solution

    2. Energy conservation: 12mpv2=e ΔV=ekQ(1r1−1r2)\tfrac12 m_p v^2=e\,\Delta V=ekQ\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right). With kQ=44.95kQ=44.95 and (10.02−10.10)=40 m−1\left(\dfrac{1}{0.02}-\dfrac{1}{0.10}\right)=40\,\text{m}^{-1}: e ΔV=(1.60×10−19)(44.95)(40)=2.88×10−16 Je\,\Delta V=(1.60\times10^{-19})(44.95)(40)=2.88\times10^{-16}\,\text{J}. Then v=2(2.88×10−16)1.67×10−27≈5.9×105 m/sv=\sqrt{\dfrac{2(2.88\times10^{-16})}{1.67\times10^{-27}}}\approx 5.9\times10^{5}\,\text{m/s}.

  3. 3. (2 pts ••) Two large parallel plates are 5.0 mm5.0\,\text{mm} apart with a potential difference of 200 V200\,\text{V} between them.

    (a) Find the magnitude of the electric field between the plates.

    (b) An electron starts from rest at the negative plate. Find its speed when it reaches the positive plate.

    Solution

    3.

    (a) E=Vd=2005.0×10−3=4.0×104 V/mE=\dfrac{V}{d}=\dfrac{200}{5.0\times10^{-3}}=4.0\times10^{4}\,\text{V/m}.

    (b) 12mev2=eV⇒v=2eVme=2(1.60×10−19)(200)9.11×10−31≈8.4×106 m/s\tfrac12 m_e v^2=eV\Rightarrow v=\sqrt{\dfrac{2eV}{m_e}}=\sqrt{\dfrac{2(1.60\times10^{-19})(200)}{9.11\times10^{-31}}}\approx 8.4\times10^{6}\,\text{m/s}.

  4. 4. (2 pts ••) A conducting sphere of radius 5.0 cm5.0\,\text{cm} carries a charge of +8.0 nC+8.0\,\text{nC}. Take V=0V=0 at infinity.

    (a) Find the potential at the surface.

    (b) Find the potential at r=15 cmr=15\,\text{cm}.

    (c) Find the potential at the center of the sphere.

    Solution

    4.

    (a) V=kQR=(8.99×109)(8.0×10−9)0.050≈1.4×103 VV=\dfrac{kQ}{R}=\dfrac{(8.99\times10^{9})(8.0\times10^{-9})}{0.050}\approx 1.4\times10^{3}\,\text{V}.

    (b) V=kQr=(8.99×109)(8.0×10−9)0.15≈4.8×102 VV=\dfrac{kQ}{r}=\dfrac{(8.99\times10^{9})(8.0\times10^{-9})}{0.15}\approx 4.8\times10^{2}\,\text{V}.

    (c) A conductor is an equipotential, so the center is at the same potential as the surface: ≈1.4×103 V\approx 1.4\times10^{3}\,\text{V}.

  5. 5. (3 pts •••) In a region of space the electric potential is V(x,y,z)=3x2y−2zV(x,y,z)=3x^2y-2z, with VV in volts and x,y,zx,y,z in meters.

    (a) Find the electric field E(x,y,z)\mathbf{E}(x,y,z) from the potential gradient.

    (b) Evaluate E\mathbf{E} at the point (2, −1, 3) m(2,\,-1,\,3)\,\text{m}.

    (c) Find the magnitude of E\mathbf{E} there.

    Solution

    5.

    (a) E=−∇V=(−∂V∂x, −∂V∂y, −∂V∂z)=(−6xy, −3x2, 2) V/m\mathbf{E}=-\nabla V=\left(-\dfrac{\partial V}{\partial x},\,-\dfrac{\partial V}{\partial y},\,-\dfrac{\partial V}{\partial z}\right)=(-6xy,\,-3x^2,\,2)\,\text{V/m}.

    (b) At (2,−1,3)(2,-1,3): E=(−6(2)(−1), −3(2)2, 2)=(12, −12, 2) V/m\mathbf{E}=(-6(2)(-1),\,-3(2)^2,\,2)=(12,\,-12,\,2)\,\text{V/m}.

    (c) ∣E∣=122+122+22=292≈17 V/m|\mathbf{E}|=\sqrt{12^2+12^2+2^2}=\sqrt{292}\approx 17\,\text{V/m}.

  6. 6. (1 pt •) An alpha particle (charge +2e+2e) is accelerated from rest through a potential difference of 1.0 kV1.0\,\text{kV}. Find its kinetic energy in joules.

    Solution

    6. KE=qV=(2)(1.60×10−19)(1.0×103)=3.2×10−16 JKE=qV=(2)(1.60\times10^{-19})(1.0\times10^{3})=3.2\times10^{-16}\,\text{J}.

  7. Extra Credit (3 pts •••)

    A thin disk of radius RR carries a uniform surface charge density σ\sigma.

    (a) By integrating the contributions of thin ring elements, find the electric potential V(x)V(x) on the disk's axis a distance xx from its center (take V=0V=0 at infinity).

    (b) From V(x)V(x), find the axial field Ex(x)E_x(x).

    (c) Show that for x≫Rx\gg R your result reduces to that of a point charge, and state what ExE_x approaches for x≪Rx\ll R.

    Solution

    Extra Credit

    (a) A ring of radius rr and width drdr carries dq=σ 2πr drdq=\sigma\,2\pi r\,dr, with every point a distance x2+r2\sqrt{x^2+r^2} from the axial point, so

    V(x)=∫0Rk σ 2πr drx2+r2=2πkσ[x2+r2]0R=σ2ε0(x2+R2−x).V(x)=\int_0^R \frac{k\,\sigma\,2\pi r\,dr}{\sqrt{x^2+r^2}}=2\pi k\sigma\Big[\sqrt{x^2+r^2}\Big]_0^R=\frac{\sigma}{2\varepsilon_0}\left(\sqrt{x^2+R^2}-x\right).

    (b) Ex=−dVdx=σ2ε0(1−xx2+R2).E_x=-\dfrac{dV}{dx}=\dfrac{\sigma}{2\varepsilon_0}\left(1-\dfrac{x}{\sqrt{x^2+R^2}}\right).

    (c) For x≫Rx\gg R, x2+R2−x≈R22x\sqrt{x^2+R^2}-x\approx \dfrac{R^2}{2x}, so V≈σR24ε0x=kQxV\approx\dfrac{\sigma R^2}{4\varepsilon_0 x}=\dfrac{kQ}{x} with Q=σπR2Q=\sigma\pi R^2 — the potential of a point charge. For x≪Rx\ll R, Ex→σ2ε0E_x\to\dfrac{\sigma}{2\varepsilon_0}, the field of an infinite charged sheet.