Question

1. consider a system in which the distance between two 0.0000025 c charges is doubled (from 0.4 m to 0.8 m). answer parts a, c, and d in full sentences.

a) predict what would happen to the force between those two charges.
b) calculate the force between the two objects for each distance.
c) did your calculations support your prediction?
d) does this finding make sense based on what you know about electrostatic forces?

1. consider a system in which the charges of two 0.0000025 c objects are doubled to 0.000005 c) and are kept at 0.4 m. answer parts a, c, and d in full sentences.

a) predict what would happen to the force between those two charges.
b) calculate the force between the two objects for each distance.
c) did your calculations support your prediction?
d) does this finding make sense based on what you know about electrostatic forces?

Explanation:

Problem 3

Part a)

According to Coulomb's Law ($F = k\frac{q_1q_2}{r^2}$), the electrostatic force is inversely proportional to the square of the distance between the charges. If the distance is doubled (from $r$ to $2r$), the force should decrease. Specifically, since $F \propto \frac{1}{r^2}$, when $r$ becomes $2r$, the new force $F'$ will be $F' = k\frac{q_1q_2}{(2r)^2}=\frac{1}{4}k\frac{q_1q_2}{r^2}=\frac{F}{4}$. So the force between the two charges should become one - fourth of its original value.

Step 1: Identify the values

We know that $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1 = q_2=2.5\times 10^{-6}\ C$, and $r = 0.4\ m$.

Step 2: Substitute into Coulomb's Law

$F_1=k\frac{q_1q_2}{r_1^2}=9\times 10^{9}\times\frac{(2.5\times 10^{-6})\times(2.5\times 10^{-6})}{(0.4)^{2}}$
First, calculate the numerator: $(2.5\times 10^{-6})\times(2.5\times 10^{-6}) = 6.25\times 10^{-12}\ C^{2}$
Then, calculate the denominator: $(0.4)^{2}=0.16\ m^{2}$
Now, $F_1 = 9\times 10^{9}\times\frac{6.25\times 10^{-12}}{0.16}$
$9\times6.25\times 10^{9 - 12}=56.25\times 10^{-3}$
$\frac{56.25\times 10^{-3}}{0.16}\approx0.3516\ N$

For $r = 0.8\ m$:

Step 1: Identify the values

$k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1 = q_2 = 2.5\times 10^{-6}\ C$, and $r = 0.8\ m$.

Step 2: Substitute into Coulomb's Law

$F_2=k\frac{q_1q_2}{r_2^2}=9\times 10^{9}\times\frac{(2.5\times 10^{-6})\times(2.5\times 10^{-6})}{(0.8)^{2}}$
The numerator is still $6.25\times 10^{-12}\ C^{2}$
The denominator: $(0.8)^{2}=0.64\ m^{2}$
$F_2=9\times 10^{9}\times\frac{6.25\times 10^{-12}}{0.64}$
$9\times6.25\times 10^{9 - 12}=56.25\times 10^{-3}$
$\frac{56.25\times 10^{-3}}{0.64}\approx0.0879\ N$ (or approximately $\frac{0.3516}{4}\approx0.0879\ N$)

Answer:

The force between the two charges will decrease to one - fourth of its original magnitude.

Part b)

We use Coulomb's Law, $F = k\frac{q_1q_2}{r^2}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1=q_2 = 0.0000025\ C=2.5\times 10^{- 6}\ C$.

For $r = 0.4\ m$: