JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A two point charges \(4 q\) and \(-q\) are fixed on the \(x-\)axis at \(x=-\frac{d}{2}\) and \(x=\frac{d}{2},\) 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
- A increase by \(\frac{2 q^{2}}{3 \pi \varepsilon_{0} d }\)
- B increase by \(\frac{3 q^{2}}{4 \pi \varepsilon_{0} d }\)
- C decrease by \(\frac{4 q^{2}}{3 \pi \varepsilon_{0} d }\)
- D decrease by \(\frac{q^{2}}{4 \pi \varepsilon_{0} d }\)
Answer & Solution
Correct Answer
(C) decrease by \(\frac{4 q^{2}}{3 \pi \varepsilon_{0} d }\)
Step-by-step Solution
Detailed explanation
Potential of \(- q\) is same as initial and final point of the path therefore potential due to \(4 q\) will only change and as potential is decreasing the energy will decrease Decrease in potential energy \(=q\left( V _{ i }- V _{ f }\right)\) Decrease in potential energy…
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