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

Q: 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

c 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 $=q\left[\frac{ k 4 q }{ d / 2}-\frac{ k 4 q }{3 d / 2}\right]=\frac{4 q ^{2}}{3 \pi \varepsilon_{0} d }$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

Aincrease by $\frac{2 q^{2}}{3 \pi \varepsilon_{0} d }$
Bincrease by $\frac{3 q^{2}}{4 \pi \varepsilon_{0} d }$
Cdecrease by $\frac{4 q^{2}}{3 \pi \varepsilon_{0} d }$
Ddecrease by $\frac{q^{2}}{4 \pi \varepsilon_{0} d }$

JEE MAIN 2020, Diffcult

STD 11 Science
NEET
STD 12 - 2. Electric potential and capacitance
Physics

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