Metal sphere 1 has a positive charge of 6.0 nC. Metal sphere...

Metal sphere 1 has a positive charge of $6.0 \mathrm{nC}$. Metal sphere
2. which is twice the diameter of sphere 1 , is initially uncharged. The spheres are then connected together by a long, thin metal wire. What are the final charges on each sphere?

Key Concepts

Electric Charge Conservation

This principle states that the total amount of electric charge in an isolated system remains constant, regardless of the redistribution of charge among its components. It is fundamental when analyzing how charge is shared between objects after they are connected or interact.

Electrostatic Equilibrium

Electrostatic equilibrium is achieved when conductors connected by a conductive path reach the same electric potential, causing charge to redistribute until no net movement occurs. This condition is essential for understanding how connected objects balance their charge.

Capacitance of Spherical Conductors

For spherical conductors, capacitance is directly related to the radius of the sphere; a larger sphere has a higher capacitance and can store more charge at a given potential. This concept is critical when determining the distribution of charge between spheres of different sizes that are interconnected.

Transcript

00:01 So we know the electric potential on the metal sphere should be equal to v, which is equal to 1 over 4 pi epsilon 0, times charge q over radius r.

00:14 And since two metalsphere will connect with each other through the metal wire, that means eventually their electric potential will be the same.

00:23 So we'll have such relationship here, which is the final potential at a metal sphere 1 should be able to the final potential at a metal sphere 2.

00:32 So now we know the energy potential on metalsphere 1 should equal to 1 over 4 pi epsilon 0 times q1f, which is the final amount charge on a sphere 1, and then over the radius on sphere 1, which is r1 here.

00:53 And this is equal to 1 over 4x00 times the final amount charge on a sphere 2 with q2 and over r2, r2 is the radius on metasphere 2.

01:06 As you can say, 1 by 4 pi epsilon 0 can be cancelled.

01:08 Well.

01:11 So therefore we have q1f over r1 is equal to q2f over r2.

01:27 And in the question, you were saying that the diameter in the metasphere 2 is twice as much the diameter of sphere 1, which means that the radius in metal sphere 2 should be twice as much the radius of the metosphere 1 as well.

01:42 Okay, so therefore we have r2 is equal to 2r1.

01:48 So if we plug in better the equation by substitute 2 r1 for r2.

01:54 We'll have q1f over r1 is equal to q2f over 2r1.

02:04 As you can tell, r1 can be canceled out.

02:09 So therefore, we have q1f, which is the final amount of charge on a sphere 1 is equal to q2f over 2...

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