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lara31 [8.8K]
3 years ago
7

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.)
Physics
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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~~~NEED HELP ASAP~~~<br>Please solve each section and show all work for each section.
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Explanation:

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass m_A as

x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)

y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)

Substituting (2) into (1), we get

\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)

where f_N= \mu_kN, the frictional force on m_A. Set this aside for now and let's look at the forces on m_B

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>

Let the x-axis be (+) up along the inclined plane. We can write the forces on m_B as

x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)

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N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)

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T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)

Substituting (7) into (4) we get

m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba

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(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag

or

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