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Electrostatic Potential and Capacitance - Potential Energy of a System of Charges

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Electrostatic Potential Energy of a system of point charges is defined as the total work done by an external agent in assembling the charges at their respective locations by bringing them from infinity.

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For a system of two charges q1q_1 and q2q_2 separated by a distance r12r_{12}, the potential energy is given by U=14πϵ0q1q2r12U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}. It is a scalar quantity and its S.I. unit is Joule (JJ).

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For a system of nn charges, the total potential energy is the algebraic sum of the potential energies of all possible pairs: U=14πϵ0∑i<jqiqjrijU = \frac{1}{4\pi\epsilon_0} \sum_{i<j} \frac{q_i q_j}{r_{ij}}.

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If a charge qq is placed in an external electric field where the potential is V(r⃗)V(\vec{r}), the potential energy of the charge is U=qV(r⃗)U = qV(\vec{r}).

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The potential energy of an electric dipole with dipole moment p⃗\vec{p} in a uniform external electric field E⃗\vec{E} is U=−p⃗⋅E⃗=−pEcos⁡θU = -\vec{p} \cdot \vec{E} = -pE \cos \theta, where θ\theta is the angle between p⃗\vec{p} and E⃗\vec{E}.

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Stable equilibrium for a dipole occurs at θ=0∘\theta = 0^\circ (Minimum Energy U=−pEU = -pE), and unstable equilibrium occurs at θ=180∘\theta = 180^\circ (Maximum Energy U=+pEU = +pE).

📐Formulae

U=14πϵ0q1q2r12U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}

Utotal=14πϵ0(q1q2r12+q2q3r23+q3q1r31)U_{total} = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right) (for three charges)

U=qV(r⃗)U = q V(\vec{r})

U=−p⃗⋅E⃗=−pEcos⁡θU = -\vec{p} \cdot \vec{E} = -pE \cos \theta

W=ΔU=pE(cos⁡θ1−cos⁡θ2)W = \Delta U = pE(\cos \theta_1 - \cos \theta_2)

💡Examples

Problem 1:

Calculate the electrostatic potential energy of a system consisting of two charges q1=7μCq_1 = 7 \mu C and q2=−2μCq_2 = -2 \mu C (and with no external field) placed at (−9 cm,0,0)(-9\text{ cm}, 0, 0) and (9 cm,0,0)(9\text{ cm}, 0, 0) respectively.

Solution:

Given: q1=7×10−6 Cq_1 = 7 \times 10^{-6} \text{ C}, q2=−2×10−6 Cq_2 = -2 \times 10^{-6} \text{ C}. Distance r=9−(−9)=18 cm=0.18 mr = 9 - (-9) = 18\text{ cm} = 0.18\text{ m}. Using the formula U=14πϵ0q1q2rU = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}: U=9×109×(7×10−6)×(−2×10−6)0.18U = \frac{9 \times 10^9 \times (7 \times 10^{-6}) \times (-2 \times 10^{-6})}{0.18} U=9×109×(−14×10−12)18×10−2U = \frac{9 \times 10^9 \times (-14 \times 10^{-12})}{18 \times 10^{-2}} U=−126×10−30.18=−0.7 JU = \frac{-126 \times 10^{-3}}{0.18} = -0.7 \text{ J}.

Explanation:

The potential energy is negative because the charges are of opposite signs, indicating an attractive force between them. The work done to assemble them is negative relative to infinity.

Problem 2:

Determine the work done to rotate an electric dipole of dipole moment 3.4×10−30 C m3.4 \times 10^{-30} \text{ C m} from its position of stable equilibrium to unstable equilibrium in a uniform electric field of 104 N/C10^4 \text{ N/C}.

Solution:

W=pE(cos⁡0∘−cos⁡180∘)W = pE(\cos 0^\circ - \cos 180^\circ) W=pE(1−(−1))=2pEW = pE(1 - (-1)) = 2pE W=2×(3.4×10−30)×104W = 2 \times (3.4 \times 10^{-30}) \times 10^4 W=6.8×10−26 JW = 6.8 \times 10^{-26} \text{ J}

Vertical calculation for the multiplier 2×342 \times 34: 34×268\begin{array}{r} 34 \\ \times 2 \\ \hline 68 \end{array}

Explanation:

Stable equilibrium is at θ1=0∘\theta_1 = 0^\circ and unstable equilibrium is at θ2=180∘\theta_2 = 180^\circ. The work done WW is the change in potential energy.