Answer :
To find the electric potential at a distance of [tex]\( 4.20 \)[/tex] meters from a charge of [tex]\( -3.37 \times 10^{-6} \)[/tex] coulombs, we use the formula for electric potential due to a point charge:
[tex]\[ V = \frac{k \cdot q}{r} \][/tex]
where:
- [tex]\( V \)[/tex] is the electric potential,
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\( 8.99 \times 10^9 \)[/tex] N m²/C²),
- [tex]\( q \)[/tex] is the charge ([tex]\( -3.37 \times 10^{-6} \)[/tex] C),
- [tex]\( r \)[/tex] is the distance from the charge (4.20 m).
Let's break down the calculation step by step:
1. Calculate the numerator [tex]\( k \cdot q \)[/tex]:
[tex]\[ k \cdot q = (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \times (-3.37 \times 10^{-6} \, \text{C}) \][/tex]
[tex]\[ k \cdot q = -3.03163 \times 10^4 \, \text{N m}^2/\text{C} \][/tex]
2. Next, divide the result by the distance [tex]\( r \)[/tex]:
[tex]\[ V = \frac{-3.03163 \times 10^4 \, \text{N m}^2/\text{C}}{4.20 \, \text{m}} \][/tex]
[tex]\[ V = -7213.4047619047615 \, \text{V} \][/tex]
The electric potential at a distance of [tex]\( 4.20 \)[/tex] meters from a [tex]\( -3.37 \times 10^{-6} \)[/tex] coulombs charge is [tex]\(-7213.4047619047615\)[/tex] volts ([tex]\( \text{V} \)[/tex]). The negative sign indicates that the potential is negative due to the negative charge.
[tex]\[ V = \frac{k \cdot q}{r} \][/tex]
where:
- [tex]\( V \)[/tex] is the electric potential,
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\( 8.99 \times 10^9 \)[/tex] N m²/C²),
- [tex]\( q \)[/tex] is the charge ([tex]\( -3.37 \times 10^{-6} \)[/tex] C),
- [tex]\( r \)[/tex] is the distance from the charge (4.20 m).
Let's break down the calculation step by step:
1. Calculate the numerator [tex]\( k \cdot q \)[/tex]:
[tex]\[ k \cdot q = (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \times (-3.37 \times 10^{-6} \, \text{C}) \][/tex]
[tex]\[ k \cdot q = -3.03163 \times 10^4 \, \text{N m}^2/\text{C} \][/tex]
2. Next, divide the result by the distance [tex]\( r \)[/tex]:
[tex]\[ V = \frac{-3.03163 \times 10^4 \, \text{N m}^2/\text{C}}{4.20 \, \text{m}} \][/tex]
[tex]\[ V = -7213.4047619047615 \, \text{V} \][/tex]
The electric potential at a distance of [tex]\( 4.20 \)[/tex] meters from a [tex]\( -3.37 \times 10^{-6} \)[/tex] coulombs charge is [tex]\(-7213.4047619047615\)[/tex] volts ([tex]\( \text{V} \)[/tex]). The negative sign indicates that the potential is negative due to the negative charge.
