Problem 38 - Fundamentals of Physics

Question

Figure 21-37 shows four identical conducting spheres that are actually well separated from one another. Sphere \(W\) (with an initial charge of zero) is touched to sphere \(A\) and then they are separated. Next, sphere \(W\) is touched to sphere \(B\) (with an initial charge of \(-32 e\) ) and then they are separated. Finally, sphere \(W\) is touched to sphere \(C\) (with an initial charge of \(+48 e\) ), and then they are separated. The final charge on sphere \(W\) is \(+18 e .\) What was the initial charge on sphere \(A ?\)

Step-by-Step Solution

Answer

The initial charge on sphere A was \(160e\).

Step 1: Charge Sharing with Sphere A

When sphere W (initially at 0 charge) touches sphere A with an unknown initial charge \( q_A \), the charges will redistribute equally as the spheres are identical. The charge on each sphere after separation will be \( \frac{q_A}{2} \), since the total initial charge \( q_A \) is shared equally between W and A.

Step 2: Charge Sharing with Sphere B

Sphere W now has a charge of \( \frac{q_A}{2} \). When W touches sphere B with an initial charge of \(-32e\), the charges will again redistribute equally. Thus, the charges on W and B after separation will be \( \frac{\frac{q_A}{2} - 32}{2} \).

Step 3: Charge Sharing with Sphere C

Sphere W now has the charge calculated from the previous step, \( \frac{\frac{q_A}{2} - 32}{2} \). When W touches sphere C (initially at \(+48e\)), the charges will redistribute equally. After separation, the charge on W becomes \( \frac{\frac{\frac{q_A}{2} - 32}{2} + 48}{2} \).

Step 4: Solving for Initial Charge on Sphere A

We now know that after touching sphere C, the final charge on W is \(+18e\). So:\[\frac{\frac{\frac{q_A}{2} - 32}{2} + 48}{2} = 18.\]Solving this equation: First clear the fractions by multiplying through by 8:\[\frac{q_A}{2} - 32 + 96 = 144.\]Simplifying, we get:\[\frac{q_A}{2} + 64 = 144.\]Subtract 64 from both sides:\[\frac{q_A}{2} = 80.\]Multiply both sides by 2 to solve for \(q_A:\)\[q_A = 160.\]

Step 5: Conclusion

The initial charge on sphere A was \(160e\).

Key Concepts

Conducting Spheres

Conducting spheres play a pivotal role in understanding charge redistribution. These spheres can be considered as perfectly conductive surfaces where charges can move freely. Conductors, by nature, allow for the easy flow of electric charge.
In our exercise, all spheres (W, A, B, and C) are identical in terms of size and material, which ensures that the charge is shared equally when they come into contact.
It's crucial to recognize that within conductors, the charge tends to move until it is uniformly distributed over the surface. This property is what allows us to calculate changes in charge accurately when the spheres touch and separate.

Initial Charge

The concept of the initial charge is foundational when analyzing problems involving charge distribution across conductive objects. Initially, the spheres start with specific charges, giving us a starting point for further calculations.
In this exercise, Sphere A has an unknown initial charge. Sphere B begins with an initial charge of \(-32 e\), and Sphere C starts with \(+48 e\). Sphere W starts with zero charge.
This initial setting allows us to trace how charges move and change during interactions between these spheres. Understanding the initial charge is vital for working out how charges redistribute after contact.

Charge Sharing

Charge sharing is the process where two conductive spheres redistribute their total charge equally between them when they come into contact. This happens because both spheres are identical in terms of capacity and geometry.
When Sphere W touches Sphere A, any total charge from both is equally divided. The same process occurs when Sphere W, now with a charge, touches Spheres B and C successively.
Every interaction leads to an even distribution, thanks to their characteristic as identical conductors. Knowing how to calculate and predict these distributions is key in solving electrostatic problems involving conductive spheres.

Final Charge Calculation

Calculating the final charge involves understanding the effect each charging and sharing step has on Sphere W. In this exercise, after several interactions of charge sharing, the final known charge on Sphere W becomes \(+18 e\).
Using the final charge of Sphere W as a clue, one can work backwards through the charge sharing steps to find out the initial unknown values.
By setting up and solving equations based on each charge sharing event as provided, we determine that the initial charge on Sphere A was \(160 e\). This process involves a sequential detailed analysis of sharing and balancing outcomes.