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Electrostatics Part 1

Section: Physics

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Question : 15 of 100

Marks: +1, -0

A two point charges

4 q

-q

are fixed on the

x

x=-\frac{d}{2}

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:

[4 Sep 2020 Shift 1]

increase by $\frac{2 q^{2}}{3 \pi \varepsilon_{0} d}$
increase by $\frac{3 q^{2}}{4 \pi \varepsilon_{0} d}$
decrease by $\frac{4 q^{2}}{3 \pi \varepsilon_{0} d}$
decrease by $\frac{q^{2}}{4 \pi \varepsilon_{0} d}$

Solution:

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(V_{i}-V_{f})

Decrease in potential energy

=q\left[ \frac{k4q}{\frac{d}{2}} - \frac{k4q}{\frac{3d}{2}} \right] = \frac{4 q^{2}}{3 \pi \varepsilon_{0} d}

Therefore correct answer is 3 .

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