 # 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: a) \frac{F}{16} b) \frac{F}{8} c) \frac{F}{4} d) \frac{F}{2} e)F Maiclubk 2021-02-21 Answered
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:
a) $\frac{F}{16}$
b) $\frac{F}{8}$
c) $\frac{F}{4}$
d) $\frac{F}{2}$
e)F
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The definition of electric force:
$F=\frac{k{q}^{2}}{{d}^{2}}$
If you change the information as stated in the problem:
$k\frac{{\left(\frac{q}{4}\right)}^{2}}{{\left(\frac{d}{2}\right)}^{2}}=\frac{4k{q}^{2}}{16{d}^{2}}=\frac{k{q}^{2}}{4{d}^{2}}=\frac{F}{4}$
###### Not exactly what you’re looking for? Jeffrey Jordon

Given

Distance between two charge is d and repulsive force is F

Solution

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 new charges are

$\frac{{q}_{1}}{4}$ and $\frac{{q}_{2}}{4}$

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

${F}_{1}=\frac{k\left(\frac{{q}_{1}}{4}×\frac{{q}_{2}}{4}\right)}{\left(\frac{d}{2}{\right)}^{2}}$

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

${F}_{1}=\frac{k{q}_{1}{q}_{2}}{4{d}^{2}}$

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