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VITEEE 2021 Paper

Section: Physics

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Question : 13 of 80

Marks: +1, -0

Two point charges +3 μC and +8 μC repel each other with a force of

40\,\text{N}

. If a charge of - 5 μC is added to each of them, then the force between them will become

$-10\,\text{N}$
$+10\,\text{N}$
$+20\,\text{N}$
$-20\,\text{N}$

Solution:

Redistribution of charges takes place.

Charge

q_{1}=3\,\mu\mathrm{C}

Charge

q_{2}=8\,\mu\mathrm{C}

When third charge

q_{3}=-5\,\mu\mathrm{C}

is added to each, then new charges on

q_{1}

and

q_{2}

will be

q_{1}'=3-5=-2\,\mu\mathrm{C}

and,

q_{2}'=8-5=3\,\mu\mathrm{C}

Now,

In first case,

40=\;\frac{1}{4\pi\varepsilon_{0}} \cdot \;\frac{3 \times 8}{r^{2}}

In second case,

F=\;\frac{1}{4\pi\varepsilon_{0}} \times \;\frac{(-2 \times 3)}{r^{2}}

\therefore \;\frac{F}{40}=\;\frac{-2 \times 3}{3 \times 8}

\;\text{ or }\;F=-10\,\text{N}

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