Two conducting spheres have identical radii. Initially they have charges of oppo

Question

Two conducting spheres have identical radii. Initially they have charges of opposite sign and unequal magnitudes with the magnitude of the positive charge larger than the magnitude of the negative charge. They attract each other with a force of 0.269 N when separated by 0.4 m. The spheres are suddenly connected by a thin conducting wire, which is then removed. Now the spheres repel each other with a force of 0.024 N. What is the magnitude of the positive charge? Answer in units of C. What is the negative charge? Answer in units of C.

Explanation / Answer

Let :F1= 0.269 N,F2= 0.024 N,r= 0.4 m,andq1>|q2

Applying Coulomb’s law, the electrostaticforce between the chargesq1andq2separatedby a
distance r is

F = Ke*q1q2/r^2

Initially both spheres have opposite charges,with q1>|q2|. from conservation of electric charges,
if the Frst sphere loses charges qx,the second sphere will gain charges qx.
After the wire is removed, both spheres have the same Fnal chargesq3, so q1qx=q3
and q2+qx=q3. from Coulomb’s law

F2 = Ke*q3^2/r^2 = q3^2 = r^2F2/Ke^2

q3 = r^2F2/Ke = (0.4)^2(0.024)/8.98*10^9N.m/C^2 = 6.539*10^-7C
Further:

-F1 =Keq1q2/r^2 = Ke*(q3+qx)(q3-qx)/r^2

= Ke (q3^2-qx^2)/r^2

-r^2F1/Ke = q3^2-qx^2
qx^2 = r^2F1/Ke^2 + r^2F2/Ke

qx = r^2(F1+F2)/Ke

= (0.4)^2(0.269+0.024)/8.987*10^9 = 2.283*10^-6C

Finally

q1 = q3+qx
= 6.539*10^-7+2.283*10^-6 = 2.9369*10^-6C

What is the negative charge?

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