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

the force on the thing and the distance the force moves the thing

wavelength divides in half and speed stays the same.