Two equal charges of magnitude Q each are placed at a distance d apart- Their electrostatic energy is E- A third charge - Q -2 is brought midway between these two charges- The electrostatic energy of the system is nowA- -2 EB- 0C- - ED- E

Μ_i = KQ^2/d = E μ_f = KQ^2/d + nbsp;K( Q)^2/d + K( Q)^2/d = KQ^2/d = E ...

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Standard XII

Physics

Potential Energy of a System of Multiple Point Charges

Two equal cha...

Question

Two equal charges of magnitude Q each are placed at a distance d apart. Their electrostatic energy is E. A third charge $- \frac{Q}{2}$ is brought midway between these two charges. The electrostatic energy of the system is now

-2E

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-E

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Solution

The correct option is B
-E

$μ_{i} = \frac{K Q^{2}}{d} = E$

$μ_{f} = \frac{K Q^{2}}{d} + \frac{K {(- Q)}^{2}}{d} + \frac{K {(- Q)}^{2}}{d}$

$= - \frac{K Q^{2}}{d} = - E$

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Two equal charges of magnitude Q each are placed at a distance d apart. Their electrostatic energy is E. A third charge $- \frac{Q}{2}$ is brought midway between these two charges. The electrostatic energy of the system is now

The correct option is B
-E

$μ_{i} = \frac{K Q^{2}}{d} = E$

$μ_{f} = \frac{K Q^{2}}{d} + \frac{K {(- Q)}^{2}}{d} + \frac{K {(- Q)}^{2}}{d}$

$= - \frac{K Q^{2}}{d} = - E$

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Two equal charges of magnitude Q each are placed at a distance d apart. Their electrostatic energy is E. A third charge −Q2 is brought midway between these two charges. The electrostatic energy of the system is now

The correct option is B -E μi=KQ2d=E μf=KQ2d+K(−Q)2d+K(−Q)2d =−KQ2d=−E

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Two equal charges of magnitude Q each are placed at a distance d apart. Their electrostatic energy is E. A third charge $- \frac{Q}{2}$ is brought midway between these two charges. The electrostatic energy of the system is now

The correct option is B
-E

$μ_{i} = \frac{K Q^{2}}{d} = E$

$μ_{f} = \frac{K Q^{2}}{d} + \frac{K {(- Q)}^{2}}{d} + \frac{K {(- Q)}^{2}}{d}$

$= - \frac{K Q^{2}}{d} = - E$