To solve the problem of finding the electric potential at [tex]\(x=0\)[/tex] due to the given charges on the [tex]\( x \)[/tex]-axis, we need to apply the formula of electric potential due to a point charge:
[tex]\[ V = \frac{kQ}{r} \][/tex]
where:
- [tex]\( V \)[/tex] is the electric potential,
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\( k = 8.99 \times 10^9 \, \mathrm{Nm^2/C^2} \)[/tex]),
- [tex]\( Q \)[/tex] is the charge,
- [tex]\( r \)[/tex] is the distance from the charge to the point where the potential is being calculated.
Given:
- [tex]\( Q_1 = +3 \, \mu C \)[/tex] at [tex]\( x_1 = 20 \, \text{cm} = 0.2 \, \text{m} \)[/tex],
- [tex]\( Q_2 = -3 \, \mu C \)[/tex] at [tex]\( x_2 = 30 \, \text{cm} = 0.3 \, \text{m} \)[/tex],
- [tex]\( Q_3 = -4 \, \mu C \)[/tex] at [tex]\( x_3 = 40 \, \text{cm} = 0.4 \, \text{m} \)[/tex].
We wish to find [tex]\( V \)[/tex] at [tex]\( x = 0 \)[/tex].
Step 1: Calculate the potential [tex]\( V_1 \)[/tex] due to [tex]\( Q_1 \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ V_1 = \frac{kQ_1}{x_1} = \frac{8.99 \times 10^9 \times 3 \times 10^{-6}}{0.2} = 134813.28 \, \text{V} \][/tex]
Step 2: Calculate the potential [tex]\( V_2 \)[/tex] due to [tex]\( Q_2 \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ V_2 = \frac{kQ_2}{x_2} = \frac{8.99 \times 10^9 \times -3 \times 10^{-6}}{0.3} = -89875.52 \, \text{V} \][/tex]
Step 3: Calculate the potential [tex]\( V_3 \)[/tex] due to [tex]\( Q_3 \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ V_3 = \frac{kQ_3}{x_3} = \frac{8.99 \times 10^9 \times -4 \times 10^{-6}}{0.4} = -89875.52 \, \text{V} \][/tex]
Step 4: Sum the potentials to find the total potential at [tex]\( x = 0 \)[/tex]:
[tex]\[ V = V_1 + V_2 + V_3 \][/tex]
[tex]\[ V = 134813.28 + (-89875.52) + (-89875.52) = -44937.76 \, \text{V} \][/tex]
Step 5: Convert the result from volts to kilovolts (1 kV = 1000 V):
[tex]\[ V = -44937.76 \, \text{V} = -44.94 \, \text{kV} \][/tex]
Since the magnitude of the electric potential at [tex]\( x = 0 \)[/tex] is [tex]\( 44.94 \, \text{kV} \)[/tex], none of the provided answer choices directly match our calculated magnitude. However, based on the magnitude of potential found, the closest answer (though incorrect as per standard rounding and potential absolute deviation) might suggest none matches directly.
Since the core calculation aligns directly with our rigorous approach, our value of [tex]\( -44.94 \, \text{kV} \)[/tex] should still be considered unless rechecked thoroughly for minor deviations as sometimes seen in selection problems allowances. Nonetheless, here the direct veracity of detailed results stands representative.
