potential is the integral of E field
integral (1/r^2) dr = -1/r + constant
so the E field drops off faster than the potential. When you move away three times you also expand your change in r to stay in proportion so the product
E delta r = delta V is the same
imagine sitting a distance r from a charge q. you measure both the electric field and electric potential. now imagine that the charge is halved and you move three time as far away. you measure the field and potential again. b what factor has the electric field changed? by what factor has the electric potential changed? are the factors the same? if so, why? If not, why not?
Electric field formula is E=KQ/r^2
Potential is: V=KQ/r
after the I move three times as far away, I got E=KQ/9r^2 and V=kQ/3r
So the Electric factor is 1/9 and the potential factor is 1/3? is that right.
if it is, then why aren't the factors the same.
2 answers
So I agree with your answer 1/9 and 1/3. I forgot to say that.