Step by step solution to "Two small charged objects repel each other with..."

High School

Maiclubk

Answered question

2021-02-21

Two small charged objects repel each other with a force F whenseparated by a distance d. if the charge on each object is reducedto one-fourth of its original value and the distance between themis reduced to d/2, the force becomes:
1) $\frac{F}{16}$
2) $\frac{F}{8}$
3) $\frac{F}{4}$
4) $\frac{F}{2}$
5)F

Answer & Explanation

2k1enyvp

Skilled2021-02-22Added 94 answers

The meaning of electric force:
$F = \frac{k q^{2}}{d^{2}}$
If we change the information as stated in the problem:
$k \frac{{(\frac{q}{4})}^{2}}{{(\frac{d}{2})}^{2}} = \frac{4 k q^{2}}{16 d^{2}} = \frac{k q^{2}}{4 d^{2}} = \frac{F}{4}$

Jeffrey Jordon

Expert2021-09-30Added 2605 answers

We have:

Distance between two charge is d and repulsive force is F

Explanation:

Force between two point charges is given by

$f = k \frac{q_{1} q_{2}}{d^{2}}$

k= coulomb's constant

d= distance between two charges

Now additional charges are

$\frac{q_{1}}{4}$ and $\frac{q_{2}}{4}$

the distance between them is $\frac{d}{2}$

$F_{1} = \frac{k (\frac{q_{1}}{4} \times \frac{q_{2}}{4})}{(\frac{d}{2})^{2}}$

$= \frac{k q_{1} q_{2}}{16 \frac{d^{2}}{4}}$

$F_{1} = \frac{k q_{1} q_{2}}{4 d^{2}}$

$F_{1} = \frac{F}{4}$

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