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.
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
So I agree with your answer 1/9 and 1/3. I forgot to say that.
Yes, you are correct in calculating the new electric field and potential values after halving the charge and moving three times as far away. The electric field factor is indeed 1/9, and the electric potential factor is 1/3.
Now, let's understand why the factors are not the same. The key difference lies in the formulas for electric field (E) and electric potential (V). In the formula for electric field (E = KQ/r^2), the distance (r) is squared in the denominator. However, in the formula for electric potential (V = KQ/r), the distance (r) is not squared. This distinction causes the factors to be different.
To analyze it further, let's consider the relationship between electric field and electric potential. Electric field is the force experienced by a unit charge placed at a given point, while electric potential is the work done per unit charge to bring a test charge from infinity to that point.
As you move three times as far away from the charge, the distance (r) in the formula for electric field becomes (3r), resulting in (E = KQ/(3r)^2 = KQ/9r^2). So, the electric field decreases by a factor of 1/9 relative to the initial field.
On the other hand, the distance (r) in the formula for electric potential becomes (3r), leading to (V = KQ/(3r) = KQ/3r). Therefore, the electric potential decreases by a factor of 1/3 relative to the initial potential.
In summary, the factors for electric field and electric potential are different because their respective formulas account for the distance in distinct ways.