Electric Potential — Question 4

Section 24-2 · Physics pages 3818–3819

Question 4 Reference changes preserve every measurable voltage difference Three nodes obey VB −VA = 20.0 V and VC −VB = −35.0 V. Initially assign VA = 100 V. x A B C Schematic; labels, not drawing lengths, specify the data. (a) Find all three potentials and VC −VA. (b) Reassign C to zero potential. Find the new node potentials and show that the work moving charge q from A to C is unchanged.
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Question 4 - Solution Reference changes preserve every measurable voltage difference x A B C (a) Method and calculation VA = 100 V, VB = 120 V, VC = 85.0 V. VC −VA = −15.0 V. (b) Interpretation and further calculation Vi′ = Vi −85.0 V : (VA′, VB′, VC′) = (15.0, 35.0, 0) V. We = q(VA −VC) = q(VA′ −VC′) = q(15.0 V). Physical check A common additive constant changes assigned energies by q times that constant but cancels from work.

Question 4

Reference changes preserve every measurable voltage difference

Three nodes obey VB−VA=20.0V_B-V_A=20.0 V and VC−VB=−35.0V_C-V_B=-35.0 V. Initially assign VA=100V_A=100 V.

Diagram shown in the original SVG worksheet.

  1. Find all three potentials and VC−VAV_C-V_A.

  2. Reassign C to zero potential. Find the new node potentials and show that the work moving charge q from A to C is unchanged.

Question 4 - Solution

Reference changes preserve every measurable voltage difference

Diagram shown in the original SVG worksheet.

(a) Method and calculation

VA=100V,VB=120V,VC=85.0V.V_A=100\,\mathrm V,\quad V_B=120\,\mathrm V,\quad V_C=85.0\,\mathrm V.VC−VA=−15.0V.V_C-V_A=-15.0\,\mathrm V.

(b) Interpretation and further calculation

Vi′=Vi−85.0V:(VA′,VB′,VC′)=(15.0,35.0,0)V.V_i\prime=V_i-85.0\,\mathrm V:\quad(V_A\prime,V_B\prime,V_C\prime)=(15.0,35.0,0)\,\mathrm V.We=q(VA−VC)=q(VA′−VC′)=q(15.0V).W_e=q(V_A-V_C)=q(V_A\prime-V_C\prime)=q(15.0\,\mathrm V).

Physical check

A common additive constant changes assigned energies by q times that constant but cancels from work.