Two Charges

• Now let us produce similar graphs (both types) for the electric potential for the case with two particles having half as much charge as before with one positioned 3cm to the left of the origin and the other 3cm right of the origin.
• The formula one enters in cell B2 is then
  =0.1/SQRT((B$1-0.03)^2+$A2^2)+0.1/SQRT((B$1+0.03)^2+$A2^2)

• Compare these graphs to the previous ones. In what ways are they similar? In what ways are they different? (Hint: close up you can "resolve" two charges; far away you cannot.)

• Now let us produce similar graphs (both types) for the electric potential for the case with four particles having a quarter as much charge as originally with one positioned 7.5cm and a second 2.5 cm to the left of the origin, and the others 2.5 cm and 7.5 cm right of the origin.
• The formula one enters in cell B2 is then
  =0.05/SQRT((B$1+0.075)^2+$A2^2)+0.05/SQRT((B$1+0.025)^2+$A2^2) +0.05/SQRT((B$1-0.025)^2+$A2^2)+0.05/SQRT((B$1-0.075)^2+$A2^2)

• In this procedure, we are beginning to see what the electric potential for a line of charges might look like.

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