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A charge of [tex][tex]$-4.33 \times 10^{-6} \, C$[/tex][/tex] is placed a certain distance from a charge of [tex][tex]$-7.81 \times 10^{-4} \, C$[/tex][/tex]. Their electric potential energy is [tex][tex]$44.9 \, J$[/tex][/tex]. How far apart are they?

(Unit: meters)


Sagot :

To determine the distance between two electrical charges based on their electric potential energy, we can use the formula for electric potential energy between two point charges:

[tex]\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \][/tex]

Where:
- [tex]\( U \)[/tex] is the electric potential energy,
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\(8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2\)[/tex]),
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the charges,
- [tex]\( r \)[/tex] is the distance between the charges.

Given the values:
- [tex]\( q_1 = -4.33 \times 10^{-6} \text{ C} \)[/tex]
- [tex]\( q_2 = -7.81 \times 10^{-4} \text{ C} \)[/tex]
- [tex]\( U = 44.9 \text{ J} \)[/tex]

We need to rearrange the formula to solve for [tex]\( r \)[/tex]:

[tex]\[ r = \frac{k \cdot q_1 \cdot q_2}{U} \][/tex]

Let's plug in the values and calculate [tex]\( r \)[/tex]:

[tex]\[ r = \frac{8.99 \times 10^9 \cdot (-4.33 \times 10^{-6}) \cdot (-7.81 \times 10^{-4})}{44.9} \][/tex]

First, calculate the numerator:

[tex]\[ \text{Numerator} = 8.99 \times 10^9 \times (-4.33 \times 10^{-6}) \times (-7.81 \times 10^{-4}) \][/tex]

[tex]\[ = 8.99 \times 10^9 \times 4.33 \times 10^{-6} \times 7.81 \times 10^{-4} \][/tex]

Since we are considering the magnitudes (both charges are negative but their product will be positive):

[tex]\[ = 8.99 \times 4.33 \times 7.81 \times 10^{9 - 6 - 4} \][/tex]

[tex]\[ = 8.99 \times 4.33 \times 7.81 \times 10^{-1} \][/tex]

[tex]\[ = 306.82657 \times 10^{-1} \][/tex]

[tex]\[ = 30.682657 \][/tex]

Now, divide by the electric potential energy [tex]\( U = 44.9 \)[/tex]:

[tex]\[ r = \frac{30.682657}{44.9} \][/tex]

[tex]\[ = 0.683456 \][/tex]

So, the distance [tex]\( r \)[/tex] between the charges is approximately:

[tex]\[ 0.6771 \, \text{meters} \][/tex]

Thus, the charges are approximately [tex]\( 0.6771 \, \text{meters} \)[/tex] apart.