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3 years ago

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How far from a -7.80 μC point charge must a 2.40 μC point charge be placed in order for the electric potential energy of the pai

r of charges to be -0.500 J ? (Take the energy to be zero when the charges are infinitely far apart.)

1 answer:

Naddik [55]3 years ago

8 0

Answer:

Distance between two point charges, r = 0.336 meters

Explanation:

Given that,

Charge 1, $q_1=-7.8\ \mu C=-7.8\times 10^{-6}\ C$

Charge 2, $q_2=2.4\ \mu C=2.4\times 10^{-6}\ C$

Electric potential energy, U = -0.5 J

The electric potential energy at a point r is given by :

$U=k\dfrac{q_1q_2}{r}$

$r=k\dfrac{q_1q_2}{U}$

$r=9\times 10^9\times \dfrac{-7.8\times 10^{-6}\times 2.4\times 10^{-6}}{-0.5}$

r = 0.336 meters

So, the distance between two point charges is 0.336 meters. Hence, this is the required solution.

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What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 43

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