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Electrostatic Potential and Capacitance - Potential Energy in an External Field

Grade 12CBSEPhysics

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

🔑Concepts

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Potential Energy of a single charge: The potential energy of a charge qq at a point with position vector r\mathbf{r} in an external electric field is given by qV(r)q V(\mathbf{r}), where V(r)V(\mathbf{r}) is the external potential at that point.

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Potential Energy of a system of two charges: In an external field, the total potential energy is the sum of the potential energies of individual charges in the external field and the mutual electrostatic potential energy between them.

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Potential Energy of a Dipole: When a dipole with dipole moment p\mathbf{p} is placed in a uniform external electric field E\mathbf{E}, it experiences a torque τ=p×E\mathbf{\tau} = \mathbf{p} \times \mathbf{E}.

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Stable and Unstable Equilibrium: A dipole is in stable equilibrium when θ=0∘\theta = 0^\circ (Potential Energy is minimum: −pE-pE) and in unstable equilibrium when θ=180∘\theta = 180^\circ (Potential Energy is maximum: +pE+pE).

📐Formulae

U=qV(r)U = q V(\mathbf{r})

U=q1V(r1)+q2V(r2)+14πϵ0q1q2r12U = q_1 V(\mathbf{r}_1) + q_2 V(\mathbf{r}_2) + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}

U(θ)=−p⋅E=−pEcos⁡θU(\theta) = -\mathbf{p} \cdot \mathbf{E} = -pE \cos \theta

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

💡Examples

Problem 1:

Two charges 7μC7 \mu\text{C} and −2μC-2 \mu\text{C} are placed at (−9 cm,0,0)(-9 \text{ cm}, 0, 0) and (9 cm,0,0)(9 \text{ cm}, 0, 0) respectively in an external electric field E=A(1/r2)E = A(1/r^2) where A=9×105 C m−2A = 9 \times 10^5 \text{ C m}^{-2}. Calculate the total electrostatic energy of the configuration.

Solution:

The potential VV for a field E=A/r2E = A/r^2 is V=A/rV = A/r. Total Energy U=q1V(r1)+q2V(r2)+kq1q2r12U = q_1 V(r_1) + q_2 V(r_2) + \frac{k q_1 q_2}{r_{12}} 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}, r1=0.09 mr_1 = 0.09 \text{ m}, r2=0.09 mr_2 = 0.09 \text{ m}, r12=0.18 mr_{12} = 0.18 \text{ m}. U=(7×10−6×9×1050.09)+(−2×10−6×9×1050.09)+9×109×7×10−6×(−2×10−6)0.18U = \left( 7 \times 10^{-6} \times \frac{9 \times 10^5}{0.09} \right) + \left( -2 \times 10^{-6} \times \frac{9 \times 10^5}{0.09} \right) + \frac{9 \times 10^9 \times 7 \times 10^{-6} \times (-2 \times 10^{-6})}{0.18} U=70−20−0.7=49.3 JU = 70 - 20 - 0.7 = 49.3 \text{ J}

Explanation:

The total energy is calculated by summing the interaction energy of each charge with the external field and the mutual interaction energy between the two charges.

Problem 2:

An electric dipole of length 2 cm2 \text{ cm} is placed with its axis making an angle of 60∘60^\circ to a uniform electric field of 105 N/C10^5 \text{ N/C}. If it experiences a torque of 83 Nm8\sqrt{3} \text{ Nm}, calculate the potential energy of the dipole.

Solution:

Torque τ=pEsin⁡θ\tau = pE \sin \theta. 83=p×105×sin⁡60∘8\sqrt{3} = p \times 10^5 \times \sin 60^\circ 83=p×105×328\sqrt{3} = p \times 10^5 \times \frac{\sqrt{3}}{2} p=16×10−5 C mp = 16 \times 10^{-5} \text{ C m} Potential Energy U=−pEcos⁡θU = -pE \cos \theta U=−(16×10−5)×105×cos⁡60∘U = -(16 \times 10^{-5}) \times 10^5 \times \cos 60^\circ U=−16×12=−8 JU = -16 \times \frac{1}{2} = -8 \text{ J}

Explanation:

First, the dipole moment pp is calculated using the torque formula, then it is substituted into the potential energy formula for a dipole in an external field.