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Chapter 17: Problem 15

A point charge \(q=+3.0\) nC moves through a potential difference $\Delta V=V_{\mathrm{f}}-V_{\mathrm{i}}=+25 \mathrm{V} .$ What is the change in the electric potential energy?

Short Answer

Expert verified
Answer: The change in electric potential energy of the point charge is 75 nJ.

Step by step solution

01

Identify the given information

The charge is \(q = 3.0 \times 10^{-9} \mathrm{C}\), and the potential difference is \(\Delta V = +25 \mathrm{V}\).
02

Use the electric potential energy formula

The formula for change in electric potential energy is \(U = q \times \Delta V\). Plug in the given values to calculate the change in electric potential energy.
03

Calculate the change in electric potential energy

Substitute the values of charge and potential difference in the formula: \(U = (3.0 \times 10^{-9} \mathrm{C}) \times (+25 \mathrm{V}) = 75 \times 10^{-9} \mathrm{J}\).
04

Convert the result to the appropriate unit

The electric potential energy is calculated in joules (J). Since the result is \(75 \times 10^{-9} \mathrm{J}\), we can convert it to nanojoules (nJ) by multiplying the value by \(10^9\). So, \(75 \times 10^{-9} \mathrm{J} = 75\) nJ.
05

Conclusion

The change in electric potential energy of the point charge is 75 nJ.

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Most popular questions from this chapter

Point \(P\) is at a potential of \(500.0 \mathrm{kV}\) and point \(S\) is at a potential of \(200.0 \mathrm{kV} .\) The space between these points is evacuated. When a charge of \(+2 e\) moves from \(P\) to \(S\), by how much does its kinetic energy change?
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