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Chapter 23: Problem 63

Two widely separated 4.0 -mm-diameter water drops each carry 15 nC. Assuming all charge resides on the drops' surfaces, find the change in electrostatic potential energy if they're brought together to form a single spherical drop.

Short Answer

Expert verified
The difference in electrostatic potential energy between the initial and final states signifies the amount of energy released or consumed in the process. This value will indicate how much energy is needed for, or released by the combination of two drops.

Step by step solution

01

Calculate the Initial Electrostatic Potential Energy

The electrostatic potential energy \( U \) between two charged particles is given by the equation \( U = k \cdot \frac{Q_1 \cdot Q_2}{r} \), where \( Q_1 \) and \( Q_2 \) are the two charges, \( r \) is the distance between them, and \( k \) is Coulomb's constant (\(8.99 \times 10^9 N \cdot m^2/C^2\)). Because the two drops are identical and both carry a charge of 15 nC (which is equivalent to \(15 \times 10^{-9} C\)), the initial energy is \( U_{initial} = k \cdot \frac{(15 \times 10^{-9})^2}{2 \cdot radius_{drop}} \). The radius of the drop can be found from the diameter which is 4.0 mm (or 4.0 \( \times 10^{-3} \) m). Half of this gives the radius(2.0 \( \times 10^{-3} \) m).
02

Calculate the Final Electrostatic Potential Energy

The final drop is a single spherical drop. The total charge of the large drop is the sum of the charges of the two smaller drops, which gives \(30 \times 10^{-9} C\). The radius of the larger drop can be found using the equation of the volume of a sphere, since the volume is conserved. This gives us \(radius_{large} = \sqrt[3]{2} \times radius_{drop}\). The energy for the larger drop is \(U_{final} = k \cdot \frac{(30 \times 10^{-9})^2}{2 \times \sqrt[3]{2} \times radius_{drop}}\).
03

Find the Change in Energy

The change in energy is given by \( \Delta U = U_{final} - U_{initial}\).

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