Question

The force f acting on a charged object varies inversely to the square of its distance r, from another charged object varies inversely to the squares of inversely to the square of its distance r, from another charged object. When two objects are 1.37 meters apart, the force acting on them is .57 newton. approximately how much force would the object feel if it were at a distance of 4.08 meters from the other object

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Answer 1

Answer: 0.064 N

Explanation:

According to Coulomb's Law:  

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$  

Where:

$F_{E}$  is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d$ is the separation distance between the charges

For the first case we have $F_{E}=0.57 N$ and $d=1.37 m$. So, we will isolate $q_{1}$ $q_{2}$ first:

$q_{1}.q_{2}=\frac{F_{E}d^{2}}{K}$  

$q_{1}.q_{2}=\frac{(0/57 N)(1.37 m)^{2}}{8.99(10)^{9} Nm^{2}/C^{2}}$  

$q_{1}.q_{2}=1.19(10)^{-10}C^{2}$  

Now, for the second case, we will have the same value for $q_{1}$ $q_{2}$, but $d=4.08 m$ this time and we have to find $F_{E}$:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{1.19(10)^{-10}C^{2}}{(4.08 m)^{2}}$  

$F_{E}=0.064 N$  As we can see this value is less than $0.57 N$ because the distance between charges was increased.