The electric potential \( V \) due to a point charge \( Q \) at a distance \( r \) is given by the formula:
\[ V = \frac{k \cdot Q}{r} \]
where:
- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 , \text{N m}^2/\text{C}^2 \),
- \( Q \) is the charge in coulombs,
- \( r \) is the distance in meters.
Given:
- \( Q = 7.5 , \mu\text{C} = 7.5 \times 10^{-6} , \text{C} \)
- \( r = 1.2 , \text{m} \)
Now, plug in the values into the formula:
\[ V = \frac{(8.99 \times 10^9) \cdot (7.5 \times 10^{-6})}{1.2} \]
Calculating the numerator:
\[ 8.99 \times 10^9 \cdot 7.5 \times 10^{-6} = 67.425 \times 10^3 = 67425 , \text{V m} \]
Now divide by \( r \):
\[ V = \frac{67425 , \text{V m}}{1.2} \approx 56187.5 , \text{V} \]
Thus,
\[ V \approx 5.619 \times 10^4 , \text{V} \]
This is approximately equal to \( 5.6 \times 10^4 , \text{V} \).
The correct answer is \( \boxed{5.6 \times 10^4 , \text{V}} \).
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