Potential Energy of a System of Charges

Electrostatic potential energy is the energy stored in a configuration of charges. It is defined as the work done by an external force in assembling the charges by bringing them from infinity to their respective locations, without acceleration.

Think of it this way: charges of the same sign repel each other. To hold them close together, you must perform work. That work doesn't vanish; it is stored in the system as potential energy.

Potential Energy of a System of Two Point Charges

Derivation

  1. Initial State: Imagine two points P1P_1 and P2P_2 at a distance r12r_{12} from each other. Initially, there are no charges in the universe (V=0V = 0 everywhere).
  2. Bringing q1q_1: We bring charge q1q_1 from infinity to point P1P_1. Since there is no electric field to oppose it, the work done W1=0W_1 = 0.
  3. The Field of q1q_1: Now, charge q1q_1 creates a potential V1V_1 at point P2P_2: V1=14πϵ0q1r12V_1 = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r_{12}}
  4. Bringing q2q_2: Now we bring charge q2q_2 from infinity to point P2P_2. The work done W2W_2 is charge ×\times potential at that point: W2=q2⋅V1=14πϵ0q1q2r12W_2 = q_2 \cdot V_1 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}
  5. Total Energy (UU): The total work done W=W1+W2W = W_1 + W_2 is stored as the potential energy UU: U=14πϵ0q1q2r12U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}

Potential Energy of a System of Three Point Charges

Derivation: Following the same logic, if we bring a third charge q3q_3 to point P3P_3:

  1. Work against q1q_1: W31=14πϵ0q1q3r13W_{31} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_3}{r_{13}}
  2. Work against q2q_2: W32=14πϵ0q2q3r23W_{32} = \frac{1}{4\pi\epsilon_0} \frac{q_2 q_3}{r_{23}}
  3. Total Energy (UU): Summing all interactions: U=14πϵ0(q1q2r12+q1q3r13+q2q3r23)U = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right) Rule: For a system of nn charges, UU is the sum of energies for all possible distinct pairs.

Potential Energy in an External Field

Sometimes, charges are placed in a region where an electric field already exists (created by some external sources we don't see).

A. Single Charge in External Field

If a charge qq is at a point with external potential V(r)V(r), the potential energy is simply: U=q⋅V(r)U = q \cdot V(r)

B. Two Charges in External Field

The total energy is the sum of their individual energies in the external field PLUS their mutual interaction energy: U=q1V(r1)+q2V(r2)+14πϵ0q1q2r12U = q_1 V(r_1) + q_2 V(r_2) + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}

Potential Energy of a Dipole in an External Field

Derivation

Consider a dipole with dipole moment p⃗\vec{p} placed in a uniform external electric field E⃗\vec{E}. The dipole experiences a torque τ=pEsin⁡θ\tau = pE \sin \theta.

electric dipole placed in a uniform external field E

  1. Work to Rotate: The small work done dWdW in rotating the dipole through a small angle dθd\theta is: dW=τdθ=pEsin⁡θdθdW = \tau d\theta = pE \sin \theta d\theta
  2. Total Work: To rotate it from angle θ0\theta_0 to θ1\theta_1: W=∫θ0θ1pEsin⁡θdθ=pE[−cos⁡θ]θ0θ1W = \int_{\theta_0}^{\theta_1} pE \sin \theta d\theta = pE [-\cos \theta]_{\theta_0}^{\theta_1} W=pE(cos⁡θ0−cos⁡θ1)W = pE (\cos \theta_0 - \cos \theta_1)
  3. Standard Reference: We choose the zero energy position at θ0=90∘\theta_0 = 90^{\circ} (where the dipole is perpendicular to the field). U(θ)=pE(cos⁡90∘−cos⁡θ)=−pEcos⁡θU(\theta) = pE (\cos 90^{\circ} - \cos \theta) = -pE \cos \theta Vector Form: U=−p⃗⋅E⃗U = -\vec{p} \cdot \vec{E}

🧠 Memory Capsule

  • Two-charge energy: U=kq1q2/rU = k q_1 q_2 / r. (Use signs of charges!).
  • Multi-charge energy: Sum of energies for all possible pairs.
  • In External Field: Utotal=∑Uext+∑UmutualU_{total} = \sum U_{ext} + \sum U_{mutual}.
  • Dipole energy: U=−pEcos⁡θU = -pE \cos \theta. Points to remember: 0∘0^\circ is stable, 180∘180^\circ is unstable.
  • Work Done in rotation: W=U(θ2)−U(θ1)=pE(cos⁡θ1−cos⁡θ2)W = U(\theta_2) - U(\theta_1) = pE(\cos \theta_1 - \cos \theta_2).

Example 1: Energy of an Equilateral Triangle

Three charges +q,+q+q, +q, and −q-q are placed at the vertices of an equilateral triangle of side aa. Find the total potential energy of the system.

Solution:

  1. Identify pairs: There are three pairs: (+q,+q)(+q,+q), (+q,−q)(+q,-q), and (+q,−q)(+q,-q).
  2. All pair separations are equal: Each distance is aa.
  3. Use pairwise energy formula: U=k(q⋅qa+q⋅(−q)a+q⋅(−q)a)U = k\left(\frac{q\cdot q}{a} + \frac{q\cdot (-q)}{a} + \frac{q\cdot (-q)}{a}\right)
  4. Simplify: U=k(q2a−q2a−q2a)=−kq2aU = k\left(\frac{q^2}{a} - \frac{q^2}{a} - \frac{q^2}{a}\right) = -k\frac{q^2}{a}
  5. Final Answer: U=−14πϵ0q2aU = -\frac{1}{4\pi\epsilon_0}\frac{q^2}{a}

Example 2: Rotating a Dipole from Aligned to Perpendicular

A dipole of moment p=5×10−8p = 5 \times 10^{-8} C m is aligned with an electric field E=104E = 10^4 N/C. How much work is required to turn the dipole so that its axis is perpendicular to the field?

Solution:

  1. Initial and final angles: θ1=0∘,θ2=90∘\theta_1 = 0^{\circ}, \quad \theta_2 = 90^{\circ}
  2. Use work formula: W=pE(cos⁡θ1−cos⁡θ2)W = pE(\cos\theta_1 - \cos\theta_2)
  3. Substitute values: W=(5×10−8)(104)(cos⁡0∘−cos⁡90∘)W = (5 \times 10^{-8})(10^4)(\cos 0^{\circ} - \cos 90^{\circ})
  4. Calculate: W=5×10−4(1−0)=5×10−4 JW = 5 \times 10^{-4}(1-0) = 5 \times 10^{-4}\ \text{J}
  5. Final Answer: W=5×10−4 JW = 5 \times 10^{-4}\ \text{J}

Example 3: Dissociating a Molecule

To pull two charges q1=1.6×10−19q_1 = 1.6 \times 10^{-19} C and q2=−1.6×10−19q_2 = -1.6 \times 10^{-19} C from an initial separation of 10−1010^{-10} m to infinity, how much external work must be done?

Solution:

  1. External work required: Wext=Uf−UiW_{\text{ext}} = U_f - U_i
  2. Final energy at infinity: Uf=0U_f = 0
  3. Initial energy: Ui=kq1q2r=9×109⋅(1.6×10−19)(−1.6×10−19)10−10U_i = k\frac{q_1 q_2}{r} = 9 \times 10^9 \cdot \frac{(1.6 \times 10^{-19})(-1.6 \times 10^{-19})}{10^{-10}}
  4. Calculate product of charges: q1q2=−2.56×10−38q_1q_2 = -2.56 \times 10^{-38}
  5. Now calculate UiU_i: Ui=9×109⋅−2.56×10−3810−10=−23.04×10−19 J=−2.304×10−18 JU_i = 9 \times 10^9 \cdot \frac{-2.56 \times 10^{-38}}{10^{-10}} = -23.04 \times 10^{-19}\ \text{J} = -2.304 \times 10^{-18}\ \text{J}
  6. External work needed: Wext=0−(−2.304×10−18)=2.304×10−18 JW_{\text{ext}} = 0 - (-2.304 \times 10^{-18}) = 2.304 \times 10^{-18}\ \text{J}
  7. Final Answer: Wext≈2.3×10−18 JW_{\text{ext}} \approx 2.3 \times 10^{-18}\ \text{J}

Example 4: Basic Two-Charge System

Two point charges +2μC+2 \mu C and −3μC-3 \mu C are separated by 10 cm. Find the potential energy of the system.

Solution:

  1. Given: q1=2×10−6 C,q2=−3×10−6 C,r=0.1 mq_1 = 2 \times 10^{-6}\ \text{C}, \quad q_2 = -3 \times 10^{-6}\ \text{C}, \quad r = 0.1\ \text{m}
  2. Use formula: U=kq1q2rU = k\frac{q_1q_2}{r}
  3. Substitute values: U=9×109⋅(2×10−6)(−3×10−6)0.1U = 9 \times 10^9 \cdot \frac{(2 \times 10^{-6})(-3 \times 10^{-6})}{0.1}
  4. Calculate: U=9×109⋅−6×10−120.1=−0.54 JU = 9 \times 10^9 \cdot \frac{-6 \times 10^{-12}}{0.1} = -0.54\ \text{J}
  5. Final Answer: U=−0.54 JU = -0.54\ \text{J}

Example 5: Energy of a Square Configuration

Four equal charges qq are placed at the corners of a square of side aa. Find the total potential energy of the system.

Solution:

  1. Count the interacting pairs: Total number of distinct pairs is 6.
  2. Side pairs: There are 4 pairs separated by distance aa. Usides=4(kq2a)U_{\text{sides}} = 4\left(k\frac{q^2}{a}\right)
  3. Diagonal pairs: There are 2 pairs separated by distance 2a\sqrt{2}a. Udiag=2(kq22a)=2 kq2aU_{\text{diag}} = 2\left(k\frac{q^2}{\sqrt{2}a}\right) = \sqrt{2}\,k\frac{q^2}{a}
  4. Add all pair energies: U=Usides+Udiag=kq2a(4+2)U = U_{\text{sides}} + U_{\text{diag}} = \frac{kq^2}{a}(4 + \sqrt{2})
  5. Final Answer: U=kq2a(4+2)U = \frac{kq^2}{a}(4 + \sqrt{2})

Example 6: Separating Two Like Charges to Infinity

How much external work is required to separate two charges q1=4μCq_1 = 4 \mu C and q2=5μCq_2 = 5 \mu C from 10 cm to infinity?

Solution:

  1. Initial energy of the system: Ui=kq1q2rU_i = k\frac{q_1q_2}{r}
  2. Substitute values: Ui=9×109⋅(4×10−6)(5×10−6)0.1U_i = 9 \times 10^9 \cdot \frac{(4 \times 10^{-6})(5 \times 10^{-6})}{0.1}
  3. Calculate: Ui=9×109⋅20×10−120.1=1.8 JU_i = 9 \times 10^9 \cdot \frac{20 \times 10^{-12}}{0.1} = 1.8\ \text{J}
  4. Final energy at infinity: Uf=0U_f = 0
  5. External work required: Wext=Uf−Ui=0−1.8=−1.8 JW_{\text{ext}} = U_f - U_i = 0 - 1.8 = -1.8\ \text{J}
  6. Interpretation: Since the charges repel, the electric field itself does positive work during separation. So the required external work in magnitude is zero if we simply let them go, but if the question asks for the change in potential energy of the system, it is −1.8-1.8 J.
  7. Final Answer: ΔU=−1.8 J\Delta U = -1.8\ \text{J}

Note: The field does +1.8 J+1.8\ \text{J} of work while the system loses 1.8 J1.8\ \text{J} of potential energy.

Example 7: External Field Potential Energy of a Single Charge

A charge of 2μC2 \mu C is placed at a point in an external field where the potential is 500 V. What is its potential energy?

Solution:

  1. Use formula: U=qVU = qV
  2. Substitute values: U=(2×10−6)(500)U = (2 \times 10^{-6})(500)
  3. Calculate: U=10−3 JU = 10^{-3}\ \text{J}
  4. Final Answer: U=1 mJU = 1\ \text{mJ}

Example 8: Rotating a Dipole from Stable to Unstable Position

An electric dipole with p=2×10−8p = 2 \times 10^{-8} C m is aligned with a field E=104E = 10^4 V/m. Find the work done in turning it from 0∘0^\circ to 180∘180^\circ.

Solution:

  1. Angles: θ1=0∘,θ2=180∘\theta_1 = 0^{\circ}, \quad \theta_2 = 180^{\circ}
  2. Use work formula: W=pE(cos⁡θ1−cos⁡θ2)W = pE(\cos\theta_1 - \cos\theta_2)
  3. Substitute: W=(2×10−8)(104)(1−(−1))W = (2 \times 10^{-8})(10^4)(1 - (-1))
  4. Calculate: W=2×10−4×2=4×10−4 JW = 2 \times 10^{-4} \times 2 = 4 \times 10^{-4}\ \text{J}
  5. Final Answer: W=4×10−4 JW = 4 \times 10^{-4}\ \text{J}

Example 9: Range of Dipole Potential Energy

For a dipole in a uniform electric field EE, what is the difference between maximum and minimum potential energy?

Solution:

  1. Minimum energy: At θ=0∘\theta = 0^{\circ}, Umin⁡=−pEU_{\min} = -pE
  2. Maximum energy: At θ=180∘\theta = 180^{\circ}, Umax⁡=+pEU_{\max} = +pE
  3. Difference: ΔU=Umax⁡−Umin⁡=pE−(−pE)=2pE\Delta U = U_{\max} - U_{\min} = pE - (-pE) = 2pE
  4. Final Answer: ΔU=2pE\Delta U = 2pE

Example 10: Work Done in Moving Along a Circle Around a Fixed Charge

A charge QQ is fixed. Another charge qq is moved along a circle of radius RR centered on QQ. Find the work done by the electrostatic force.

Solution:

  1. Potential energy at any point on the circle: U=kQqRU = k\frac{Qq}{R}
  2. Since the radius remains constant, the separation between the charges does not change.
  3. Therefore potential energy remains constant throughout the motion: ΔU=0\Delta U = 0
  4. Work done by electrostatic force is: W=−ΔU=0W = -\Delta U = 0
  5. Final Answer: W=0W = 0

Example 11: Potential Energy of Two Charges in an External Field

Two charges q1=1μCq_1 = 1 \mu C and q2=−2μCq_2 = -2 \mu C are placed at points where the external potentials are 300 V and 100 V respectively. Their separation is 0.2 m. Find the total potential energy of the system.

Solution:

  1. Use the total energy formula: U=q1V(r1)+q2V(r2)+kq1q2r12U = q_1V(r_1) + q_2V(r_2) + k\frac{q_1q_2}{r_{12}}
  2. External-field contribution: q1V(r1)+q2V(r2)=(1×10−6)(300)+(−2×10−6)(100)q_1V(r_1) + q_2V(r_2) = (1 \times 10^{-6})(300) + (-2 \times 10^{-6})(100) =3×10−4−2×10−4=1×10−4 J= 3 \times 10^{-4} - 2 \times 10^{-4} = 1 \times 10^{-4}\ \text{J}
  3. Mutual interaction term: kq1q2r12=9×109⋅(1×10−6)(−2×10−6)0.2k\frac{q_1q_2}{r_{12}} = 9 \times 10^9 \cdot \frac{(1 \times 10^{-6})(-2 \times 10^{-6})}{0.2} =9×109⋅−2×10−120.2=−9×10−2 J= 9 \times 10^9 \cdot \frac{-2 \times 10^{-12}}{0.2} = -9 \times 10^{-2}\ \text{J}
  4. Total energy: U=1×10−4−9×10−2=−8.99×10−2 JU = 1 \times 10^{-4} - 9 \times 10^{-2} = -8.99 \times 10^{-2}\ \text{J}
  5. Final Answer: U=−8.99×10−2 JU = -8.99 \times 10^{-2}\ \text{J}