What must be the distance between two equal and opposite point charges (say $+ q$ and $- q$ ) for the electrostatic force between them to have a magnitude of $16 N$ ?

4 \sqrt{k g} m e t r e

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\frac{q}{4} \sqrt{k} m e t r e

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4 k q m e t e r

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\frac{4 k}{q} m e t r e

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Solution

The correct option is B $\frac{q}{4} \sqrt{k} m e t r e$
Electrostatic force between two equal and opposite charges is $16 N$ .
$F = \frac{k q (- q)}{r^{2}}$

$∣ ∣ ∣ \frac{- k q^{2}}{r^{2}} ∣ ∣ ∣ = 16 N$

$r = \sqrt{\frac{k q^{2}}{16}}$
$r = \frac{q}{4} \sqrt{k}$ meters

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What must be the distance between two equal and opposite point charges (say +q and −q) for the electrostatic force between them to have a magnitude of 16 N?

The correct option is B q4√k metreElectrostatic force between two equal and opposite charges is 16 N. F=k q(−q)r2∣∣∣−k q2r2∣∣∣=16 Nr=√k q216 r=q4√k meters

What must be the distance between two equal and opposite point charges (say $+ q$ and $- q$ ) for the electrostatic force between them to have a magnitude of $16 N$ ?

The correct option is B $\frac{q}{4} \sqrt{k} m e t r e$
Electrostatic force between two equal and opposite charges is $16 N$ .
$F = \frac{k q (- q)}{r^{2}}$

$∣ ∣ ∣ \frac{- k q^{2}}{r^{2}} ∣ ∣ ∣ = 16 N$

$r = \sqrt{\frac{k q^{2}}{16}}$
$r = \frac{q}{4} \sqrt{k}$ meters