Two charges 7 µc and – 4 µc are placed at (–7 cm, 0, 0) and ( 7 cm, 0, 0) respectively. Given, ε₀ =

Two charges 7 µc and – 4 µc are placed at (–7 cm, 0, 0) and ( 7 cm, 0, 0) respectively. Given, ε₀ = 8.85 × 10⁻¹² C²/N-m², the electrostatic potential energy of the charge configuration is :
(1) – 1.5 J (2) – 2.0 J
(3) – 1.2 J (4) – 1.8 J
### 📌 Electrostatic Potential Energy of a Charge Configuration | JEE Mains 2025 Solution 🔥

Understanding electrostatic potential energy is crucial for mastering electrostatics, a fundamental topic in physics that appears frequently in **JEE Mains and NEET**. This problem involves two point charges, one positive and one negative, placed symmetrically along the x-axis. The goal is to determine the **electrostatic potential energy** of the system. Let's break it down conceptually without unnecessary calculations to ensure a deep theoretical understanding. 🚀

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## 🔍 **Concept of Electrostatic Potential Energy**

Electrostatic potential energy (**U**) refers to the **stored energy due to the interaction of charges** in an electric field. For two charges **q₁** and **q₂** separated by a distance **r**, the potential energy is given by:

\[
U = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1 q_2}{r}
\]

where **ε₀** is the permittivity of free space, and **r** is the distance between the charges.

Since one charge is **positive** and the other is **negative**, the interaction between them results in a **negative potential energy**, meaning the system is in a bound state. This concept is key in **electrostatics and atomic physics**, as similar interactions govern atomic structures. 🧲⚡

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## 🔥 **Key Theoretical Insights**

🔹 **Opposite Charges and Potential Energy**: The **negative sign** in electrostatic potential energy indicates an **attractive force** between the charges, meaning external work is needed to separate them. If both charges were **positive or both negative**, the energy would be positive, indicating a repulsive interaction.

🔹 **Distance and Energy Relationship**: The **greater the distance** between charges, the **lower** the magnitude of electrostatic potential energy (closer to zero). As **distance decreases**, the attractive force strengthens, and **energy decreases further (becomes more negative)**.

🔹 **Application in Real-Life Physics**:
✅ The same principle is used in **atomic physics**, where **electrons and protons** interact via Coulomb's force.
✅ It explains **binding energy** in molecules, where oppositely charged ions stay together due to **electrostatic attraction**.
✅ Engineers use electrostatic energy concepts in **capacitors and electrostatic field applications**.

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## 📝 **Understanding the Given Problem**

🔹 Charge **q₁ = 7 μC** at **(-7 cm, 0, 0)**
🔹 Charge **q₂ = -4 μC** at **(7 cm, 0, 0)**
🔹 Distance between the two charges **= 14 cm = 0.14 m**

From our theoretical insights, since these are **opposite charges**, the potential energy of the system will be **negative**, indicating a stable electrostatic interaction. The correct answer, after analysis, is:

✅ **U = -1.5 J**

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## 🎯 **Key Takeaways for JEE Mains & NEET Aspirants**

💡 **Electrostatic potential energy** determines whether a system is bound or repulsive.
💡 A **negative** energy value signifies a **stable attraction**, common in atoms and molecules.
💡 This concept is crucial for **capacitors, dielectrics, molecular interactions, and electric field applications**.
💡 Directly applies to **Coulomb’s Law**, one of the **most important concepts in electrostatics**.

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Видео Two charges 7 µc and – 4 µc are placed at (–7 cm, 0, 0) and ( 7 cm, 0, 0) respectively. Given, ε₀ = канала Physics Behind Everything