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What happens to electrostatic potential ...

What happens to electrostatic potential energy of a two electron system, if one electrons brought towards another electron?

A

It become zero

B

It decreases

C

It increases

D

It remains same

Text Solution

AI Generated Solution

The correct Answer is:
To determine what happens to the electrostatic potential energy of a two-electron system when one electron is brought closer to another electron, we can follow these steps: ### Step 1: Understand the Electrostatic Potential Energy Formula The electrostatic potential energy (U) between two point charges (Q1 and Q2) separated by a distance (r) is given by the formula: \[ U = k \frac{Q_1 Q_2}{r} \] where \( k \) is Coulomb's constant. ### Step 2: Identify the Charges In our case, we have two electrons. The charge of an electron is negative, specifically \( Q_1 = Q_2 = -e \) (where \( e \) is the magnitude of the charge of an electron). ### Step 3: Write the Initial Potential Energy When two electrons are at a distance \( r \) apart, the initial potential energy (U_initial) can be expressed as: \[ U_{initial} = k \frac{(-e)(-e)}{r} = k \frac{e^2}{r} \] This value is positive because both charges are negative, and the product of two negative charges is positive. ### Step 4: Consider Bringing One Electron Closer Now, if we bring one electron closer to the other, the distance between them decreases. Let’s say we bring the second electron to a distance \( r' \) (where \( r' < r \)). ### Step 5: Write the New Potential Energy The new potential energy (U_new) when the distance is \( r' \) becomes: \[ U_{new} = k \frac{(-e)(-e)}{r'} = k \frac{e^2}{r'} \] Since \( r' < r \), it follows that \( \frac{1}{r'} > \frac{1}{r} \), which means: \[ U_{new} > U_{initial} \] ### Step 6: Conclusion As a result, when one electron is brought closer to another electron, the electrostatic potential energy of the system increases. This is because like charges repel each other, and as they get closer, the potential energy associated with their positions increases. ### Final Answer The electrostatic potential energy of a two-electron system increases when one electron is brought closer to the other. ---
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Knowledge Check

  • Potential energy of electron present in He^(+) is:

    A
    `(e^(2))/(2pi epsilon_(0)r)`
    B
    `(3e^(2))/(4pi epsilon_(0)r)`
    C
    `(-2e^(2))/(4pi epsilon_(0)r)`
    D
    `(-e^(2))/(4piepsilon_(0)r^(2))`
  • In above Q., the potential energy of the electron is:

    A
    `-1.7 eV`
    B
    `-3.4 eV`
    C
    `-6.8 eV`
    D
    `-13.4 eV`
  • The potential energy of an electron in the fifth orbit of hydrogen atom is

    A
    `0.54 eV`
    B
    `- 0.54 eV`
    C
    `1.08 eV`
    D
    `- 1.08 eV`
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