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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]-7.81 \times 10^{-4} C[/tex]. Their electric potential energy is [tex]44.9 J[/tex]. How far apart are they?

(Unit: m)


Sagot :

To solve for the distance between the two charges given their electric potential energy, we can use the formula for electric potential energy between two point charges:

[tex]\[ U = k \frac{|q_1 \cdot q_2|}{r} \][/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 magnitudes of the charges,
- [tex]\( r \)[/tex] is the distance between the charges.

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

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

Given:
- [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]

and

[tex]\[ k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \][/tex]

First, calculate the absolute value of the product of the charges:

[tex]\[ |q_1 \cdot q_2| = |(4.33 \times 10^{-6} \, \text{C}) \cdot (-7.81 \times 10^{-4} \, \text{C})| \][/tex]

[tex]\[ = |4.33 \times 10^{-6} \cdot (-7.81 \times 10^{-4})| \][/tex]

[tex]\[ = |4.33 \cdot -7.81| \times 10^{-10} \][/tex]

[tex]\[ = 33.8073 \times 10^{-10} \][/tex]

[tex]\[ = 3.38073 \times 10^{-9} \, \text{C}^2 \][/tex]

Next, plug this value into the rearranged formula along with the given values for [tex]\( k \)[/tex] and [tex]\( U \)[/tex]:

[tex]\[ r = 8.99 \times 10^9 \frac{\text{N m}^2}{\text{C}^2} \times \frac{3.38073 \times 10^{-9} \, \text{C}^2}{44.9 \, \text{J}} \][/tex]

[tex]\[ = \frac{8.99 \times 10^9 \times 3.38073 \times 10^{-9}}{44.9} \][/tex]

[tex]\[ = \frac{30.4057627}{44.9} \][/tex]

[tex]\[ = 0.6770991692650334 \, \text{m} \][/tex]

Therefore, the distance between the two charges is approximately [tex]\( 0.677 \)[/tex] meters.