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Natalija [7]
3 years ago
10

The force of attraction between two like charged table tennis balls is 2.4 × 10-5 newtons. If the charge on the one is 3.8 × 10-

8 coulombs and on the other is 3.0 × 10-8 coulombs, what is the distance between the two charges? (k = 9.0 × 109 newton·meter2/coulombs2) A. 0.11 meters B. 0.24 meters C. 0.45 meters D. 0.65 meters
Physics
2 answers:
k0ka [10]3 years ago
7 0
To have a force of 3.8x10-8 the two chaarges must be 0.65m apart. 

Radda [10]3 years ago
3 0
<span>F = k*Q1*Q2 / r²  (Coulombs law)

Now we just put in the values: 2.4*10^{-5} = 9*10^{9}*\frac{ 3.0*10^{-8} * 3.8*10^{-8} }{ r² }\]

Solving for r:

r = squareroot{ 9*10^{9}*/ 3.0*10^{-8} * 3.8*10^{-8} }{ 2.5*10^{-5}

Which gives r = 0653835. The answer is D.</span>
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Read 2 more answers
Two spheres have identical charges and are 75 cm apart. the force between them is +0.30 n. what is the magnitude of the charge o
ella [17]

Answer:

4.33\cdot 10^{-6}C, charges are both positive or both negative

Explanation:

The electrostatic force between the two spheres is given by

F=k\frac{q_1 q_2}{r^2}

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k is the Coulomb's constant

q1 and q2 are the charges on the two spheres

r is the distance between the centres of the two spheres

In this problem, we have

F=+0.30 N is the force

r=75 cm=0.75 m is the distance between the spheres

q_1 =q_2 =q because the two spheres have identical charge

Solving the formula for q, we find

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(+0.30 N)(0.75 m)^2}{9\cdot 10^9}}=4.33\cdot 10^{-6}C

And the two charges have the same sign (so, both positive or both negative), since the sign of the force is positive (+0.30 N), so it is a repulsive force.

5 0
3 years ago
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