<p><a name=\"11\">&#160;Two identical conducting spheres, fixed in place, attrac

Question

<p><a name="11">&#160;Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.3021&#160;<em>N</em>when separated by 50&#160;<em>cm</em>, center-to-center. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.1902&#160;<em>N</em>. What were the initial charges on the spheres? Since one is negative and you cannot tell which is positive or negative, there are two solutions. Take the absolute value of the charges and enter the smaller value here.&#160;</a></p>
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<p>I cannot figure out the smaller value. My answer I keep getting is incorrect. However, the larger value is 6.00 X 10 -6 C.</p>

Explanation / Answer

After connected by the wire, two spheres now have the same magnitude of charges since the two spheres have identical size.
Let Q be the charge in both sphere
Repel force = kQ2/r2 = 0.1902N (assume that both spheres have positive charges)

Solve for Q=2.3e-06C

Therfore, total charge = 2*2.3e-06 = 4.6e-06 C

Before connecting the wire, both sphere have different charges but the total of charges in both spheres must equal to 4.6e-06 C

Let A be number of charges in a sphere

Attraction force = k(A)(4.6e-06 - A) / r2 = 0.302N

plug in k and r and solve quadratic equation for A, A = -1.4e-06 C

Therefore, charges in the other sphere = 4.6e-06 - (-1.4e-06) = 6e-06C

Smaller value = 1.4e-06 C here

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