Points A, B, C, and D are at the corners of a square area in an electric field, with B adjacent to A and C diagonally across from A. The potential difference between A and C is the negative of that between A and B and the same as that between B and D. Part B What is the potential difference between C and D? Delta V_CD = Delta V_AB Delta V_CD = -Delta V_AB Delta V_CD = -2 Delta V_AB Delta V_CD = 0 Part C What is the potential difference between A and D? Delta V_CAD = Delta V_AB Delta V_AD = -2 Delta V_AB Delta V_AD = Delta V_AB Delta V_AD = 0

b)    V_CD = - V_AB ,  c)      V_AD = -2 V_AB

Explanation:

b) In this exercise of potential differences they indicate that the diagonal potential is the negative of the leg potential

      V_AC = - V_AB

     V_BD = - V_CD

Thus

         V_CD = V_AC

Finally

           V_CD = - V_AB

c) The potential is a scalar quantity so it can be added algebraically, they give us the condition

          V_AC = -V_AB

          V_AB + V_BC = - V_AB

          V_BC = - 2 V_AB

For the other triangle

          V_BD = V_AC

          V_AB + V_AD =  V_AC

We replace

          V_AB + V_AD =  - V_AB

          V_AD = -2 V_AB

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explanation:

i'm only doing this the points

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i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

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