Edits to “Statement”, “Solution”, “Answer”

Tete edited Aug 27, 2026, 4:47 PM

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revision #20371 parent #20370 ← older

en/6.5.24.md +3 −3

		@@ -1,11 +1,11 @@
			### Statement

		−	$6.5.24.$ [Insert the problem statement]
		+	$6.5.24.$ A uniformly charged sheet having the shape of an isosceles right triangle was folded in half. In this case, work $A$ was performed against the forces of the electric field. What kind of work do you need to do to fold the resulting triangle again in the same way?

			### Solution

		−	By folding an isosceles right triangle in half, its new linear dimension is $1/\sqrt2$ of its original dimension. If the electrical potential energy of the original charged triangle is $U$, then the potential energy of the new triangle is $\sqrt2U$, because potential energy varies inversely with distance ~~while th~~e total charge remains the same. Thus, the work done in the first folding is $A=\sqrt2U-U$. If the new triangle is folded once more, then the potential energy of the last triangle is $(\sqrt2)^2U$, and the work done in the second folding is $(\sqrt2)^U-\sqrt2U=\sqrt2A$.
		+	By folding an isosceles right triangle in half, its new linear dimension is $1/\sqrt2$ of its original dimension. If the electrical potential energy of the original charged triangle is $U$, then the potential energy of the new triangle is $\sqrt2\,U$, because potential energy varies inversely with distance between charges while total charges remain the same. Thus, the work done in the first folding is $A=\sqrt2\,U-U$. If the new triangle is folded once more, then the potential energy of the last triangle is $(\sqrt2)^2\,U$, and the work done in the second folding is $(\sqrt2)^2\,U-\sqrt2\,U=\sqrt2A$.

			#### Answer

		−	[Insert a concise answer or boxed result]
		+	$\sqrt2A$

	@@ -1,11 +1,11 @@
	### Statement		### Statement

	$6.5.24.$ [Insert the problem statement]		$6.5.24.$ A uniformly charged sheet having the shape of an isosceles right triangle was folded in half. In this case, work $A$ was performed against the forces of the electric field. What kind of work do you need to do to fold the resulting triangle again in the same way?

	### Solution		### Solution

	By folding an isosceles right triangle in half, its new linear dimension is $1/\sqrt2$ of its original dimension. If the electrical potential energy of the original charged triangle is $U$, then the potential energy of the new triangle is $\sqrt2U$, because potential energy varies inversely with distance ~~while th~~e total charge remains the same. Thus, the work done in the first folding is $A=\sqrt2U-U$. If the new triangle is folded once more, then the potential energy of the last triangle is $(\sqrt2)^2U$, and the work done in the second folding is $(\sqrt2)^U-\sqrt2U=\sqrt2A$.		By folding an isosceles right triangle in half, its new linear dimension is $1/\sqrt2$ of its original dimension. If the electrical potential energy of the original charged triangle is $U$, then the potential energy of the new triangle is $\sqrt2\,U$, because potential energy varies inversely with distance between charges while total charges remain the same. Thus, the work done in the first folding is $A=\sqrt2\,U-U$. If the new triangle is folded once more, then the potential energy of the last triangle is $(\sqrt2)^2\,U$, and the work done in the second folding is $(\sqrt2)^2\,U-\sqrt2\,U=\sqrt2A$.

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		$\sqrt2A$