Question

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Answer 1

QUESTION:

Two charges +4q and +q are separated by distance r, where should a third point charge Q be placed so that the whole system remains in equilibrium.

ANSWER:

opton d) r/3 from +q charge

GIVEN:

$Q_1 = +4q$

$Q_2 = + q$

Distance = r

TO FIND:

The value of x

FORMULAE:

$F = \frac{1}{4 \pi \epsilon_o} \times \frac{Q_1Q_2}{ {r}^{2} }$

$F_1 = F_2$

EXPLANATION:

$F_1 = 9 \times {10}^{9} \left( \frac{4qQ}{ {x}^{2} } \right) \\ \\ F_2 =9\times {10}^{9}(\frac{qQ}{{(r - x)}^{2}}\\ \\ 9 \times {10}^{9} \left( \frac{4qQ}{ {x}^{2} } \right) =9 \times {10}^{9} \left( \frac{qQ}{ {(r - x)}^{2}}\right) \\ \\\frac{4qQ}{ {x}^{2} } = \frac{qQ}{ {(r - x)}^{2}} \\ \\ \frac{4}{ {x}^{2} } = \frac{1}{ {(r - x)}^{2} } \\ \\ \textsf{\bf{Take square root on both sides}}\\ \\ \frac{2}{x} = \frac{1}{r - x} \\ \\ 2r - 2x =x \\ \\ 3x = 2r \\ \\ x = \frac{2}{3} r$

2r/3 from +4q charge as it is not given we will subtract it from r

r - 2/3r = r/3

HENCE THE CHARGE Q SHOULD BE PLACED AT A DISTANCE OF r/3 FROM +q CHARGE.

NOTE: REFER ATTACHMENT FOR DIAGRAM.