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Physics NCERT Exemplar Solutions Class 11th Chapter Ten Electrostatics Physics Electrostatic Potential and Capacitance Physics Class 12th

A two point charges $4 q$ and $- q$ are fixed on the $x$ -axis at $x = - \frac{d}{2}$ $\frac{}{}$ and $x = \frac{d}{2}$ $\frac{}{}$ , respectively. If a third point charge ' $q$ ' is taken from the origin to $x = d$ along the semicircle as shown in the figure, the energy of the charge will:

Option 1 -

Decrease by $\frac{^{}}{_{}}$ $\frac{4 q^{2}}{3 π \in_{0} d}$

Option 2 -

Increase by $\frac{^{}}{_{}}$ $\frac{2 q^{2}}{3 π \in_{0} d}$

Option 3 -

Increase by $\frac{^{}}{_{}}$ $\frac{3 q^{2}}{4 π \in_{0} d}$

Option 4 -

Decrease by $\frac{q^{2}}{4 π \in_{0} d}$ $\frac{^{}}{_{}}$

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Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

$U_{initial} = \frac{k (4 q) (q)}{(d / 2)} + \frac{k (q) (- q)}{(d / 2)}$

$\begin{array}{r} \Rightarrow \frac{6 k q^{2}}{d} \\ \Rightarrow U_{final} = \frac{4 (4 q) (q)}{(\frac{3 d}{2})} + \frac{k (q) (- q)}{(d / 2)} \\ \Rightarrow \frac{2}{3} \frac{k q^{2}}{d} \\ \Rightarrow Δ U = (\frac{2}{3} - 6) \frac{k q^{2}}{d} \Rightarrow \frac{- 16}{3} \frac{{k q}^{2}}{d} \end{array}$

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A two point charges $4 q$ and $- q$ are fixed on the $x$ -axis at $x = - \frac{d}{2}$ $\frac{}{}$ and $x = \frac{d}{2}$ $\frac{}{}$ , respectively. If a third point charge ' $q$ ' is taken from the origin to $x = d$ along the semicircle as shown in the figure, the energy of the charge will:

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A two point charges 4 q and - q are fixed on the x -axis at x = - d 2 and x = d 2 , respectively. If a third point charge ' q ' is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will:

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A two point charges $4 q$ and $- q$ are fixed on the $x$ -axis at $x = - \frac{d}{2}$ $\frac{}{}$ and $x = \frac{d}{2}$ $\frac{}{}$ , respectively. If a third point charge ' $q$ ' is taken from the origin to $x = d$ along the semicircle as shown in the figure, the energy of the charge will: