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3 years ago

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Two charged particles attract each other with a force of magnitude F acting on each. If the charge of one is doubled and the dis

tance separating the particles is also doubled, the force acting on each of the two particles has magnitude
(a) F/2,
(b) F/4,
(c) F,
(d) 2F,
(e) 4F,
(f) None of the above.

1 answer:

irakobra [83]3 years ago

4 0

Answer:

F/2

Explanation:

In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant

Hence;

F= KQ1Q2/r^2 ------(1)

Where the charge on Q1 is doubled and the distance separating the charges is also doubled;

F= K2Q1 Q2/(2r)^2

F2= 2KQ1Q2/4r^2 ----(2)

F2= F/2

Comparing (1) and (2)

The magnitude of force acting on each of the two particles is;

F= F/2

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4 years ago

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Answer:

$q_1=9.9998\mu C$ and $q_2=0.0002\mu C$

Or

$q_1=0.00016\mu C$ and $q_2=9.99984\mu C$

Explanation:

We are given that

Force between two charges=0.036 N= $36\times 10^{-3}N$

Distance between two charges, r=2cm= $2\times 10^{-2}$ m

1m=100cm

Sum of two charges= $10\mu C$

Let one charge= $q_1=q\mu C=q\times 10^{-6}C$

$q_2=(10-q)\times 10^{-6} C$

We know that

Electric force between two charges

$F=\frac{kq_1q_2}{r^2}$

Where $k=\frac{1}{4\pi \epsilon_0}=9\times 10^{9}$

Using the formula

$36\times 10^{-3}=9\times 10^{9}\times \frac{q\times 10^{-6}\times(10-q)\times 10^{-6}}{(2\times 10^{-2})^2}$

$\frac{144\times 10^{-7}}{9\times 10^{9}\times 10^{-12}}=q(10-q)$

$0.0016=10q-q^2$

$q^2-10q+0.0016=0$

$10000q^2-100000q+16=0$

$q=\frac{100000\pm\sqrt{(100000)^2-4\times 10000\times 16}}{2\times 10000}$

Using the formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$q=9.999$ and $q=0.00016$

$q_2=10-9.9998=0.0002$

$q_2=10-0.00016=9.99984$

Hence, two charges are

$q_1=9.9998\mu C$ and $q_2=0.0002\mu C$

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$q_1=0.00016\mu C$ and $q_2=9.99984\mu C$

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Answer:

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Explanation:

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The intensity of the unpolarised light $I_o$

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Generally the intensity of light emerging from the first polarizer is mathematically represented as

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