Two identical point charges are separated by a distance r. The electric potential at the midpoint between them is 210 V. What is the electric potential at a point 25% of the way from one charge to the other along the line joining them?

• A. 110 V
• B. 210 V
• C. 440 V
• D. 880 V

Answer: (C)

Electric potential from multiple charges adds up. For two identical point charges q separated by distance r, the potential at a point is V = kq/d1 + kq/d2, where d1 and d2 are the distances to each charge.

At the midpoint, both distances are r/2, so V_mid = kq/(r/2) + kq/(r/2) = 4kq/r. Since V_mid is 210 V, we have kq/r = 52.5 V.

Now consider a point 25% of the way from one charge toward the other. The distances to the charges are r/4 and 3r/4. The potential there is V = kq/(r/4) + kq/(3r/4) = (kq/r)(4 + 4/3) = (kq/r) * 16/3 = 52.5 * 16/3 = 280 V.

So the potential at that point is 280 V. The listed options don’t include this value, indicating a mismatch in the options.