Question

Problem Two equal like charges in air repel
achother with a force F. By what percentage should each charge
reduced so that the force between them in a medium of dielectric
bastant 2 reduces by 28% ?

Plz solve it class 12​

Answer 1

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Answer 2

Answer:

The change in magnitude of the charge is 20%.

Explanation:

Given are 2 charges of equal magnitude and same polarity in air which repel each other with a force of magnitude F.

$F=\frac{Q^{2} }{4\pi E_{0}r^{2} }$ .........(1)

Now when the system is transferred in a medium of dielectric constant 2

and the force is reduced by 28% (New force=0.72F)

Then let the new charge be Q',

$0.72F=\frac{Q'^{2} }{4\pi 2E_{0}r^{2} }$........(2)

Dividing equations 1 and 2,

$\frac{Q'^{2} }{Q^{2} } =1.44\\\frac{Q'}{Q}=1.2$

Q'=1.2Q

So the percentage change in the charge is,

$\frac{Q'-Q}{Q}\times 100\\\frac{0.2}{1} \times 100=20 \%$

Hence the change in the charge is 20%.