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title: "Two point charges with charge magnitudes \\(+q\\) and \\(+3q\\) are fixed a distance \\(d\\) apart. The electric potential energy of this two-charge system is \\(U_0\\). The charge \\(+q\\) is then replaced by a charge \\(+2q\\), and the distance between the two charges is reduced to \\(\\dfrac{d}{2}\\). What is the new electric potential energy of the system in terms of \\(U_0\\)?"
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url: "https://nerd-notes.com/ubq/118035/"
date_modified: "2026-08-04T08:04:34+00:00"
---

# Two point charges with charge magnitudes \(+q\) and \(+3q\) are fixed a distance \(d\) apart. The electric potential energy of this two-charge system is \(U_0\). The charge \(+q\) is then replaced by a charge \(+2q\), and the distance between the two charges is reduced to \(\dfrac{d}{2}\). What is the new electric potential energy of the system in terms of \(U_0\)?

Two point charges with charge magnitudes \(+q\) and \(+3q\) are fixed a distance \(d\) apart. The electric potential energy of this two-charge system is \(U_0\). The charge \(+q\) is then replaced by a charge \(+2q\), and the distance between the two charges is reduced to \(\dfrac{d}{2}\). What is the new electric potential energy of the system in terms of \(U_0\)?

- **A.** \(4U_0\)
- **B.** \(2U_0\)
- **C.** \(U_0\)
- **D.** \(\dfrac{1}{2}U_0\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118035/*