Q181 / 200 · MHT CET - Medical 2026 (24 April Shift 2) · Physics
Three point charges $q_1 = 1\ \mu\text{C}$, $q_2 = -2\ \mu\text{C}$, and $q_3 = 3\ \mu\text{C}$ are placed at $(1, 0, 0)$, $(0, 2, 0)$ and $(0, 0, 3)$ respectively. The electric potential at origin is $\left\{\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9\ \text{SI unit}\right\}$
- A$10^6\ \text{V}$
- B$3 \times 10^3\ \text{V}$
- C$6 \times 10^6\ \text{V}$
- Correct answerD$9 \times 10^3\ \text{V}$
Solution
The electric potential at the origin is given by the superposition principle:
$V = \dfrac{1}{4\pi\epsilon_0} \left( \dfrac{q_1}{r_1} + \dfrac{q_2}{r_2} + \dfrac{q_3}{r_3} \right)$
The distances of the charges from the origin are:
$r_1 = \sqrt{1^2 + 0^2 + 0^2} = 1\ \text{m}$
$r_2 = \sqrt{0^2 + 2^2 + 0^2} = 2\ \text{m}$
$r_3 = \sqrt{0^2 + 0^2 + 3^2} = 3\ \text{m}$
Substituting the values of charges and distances:
$V = 9 \times 10^9 \left( \dfrac{1 \times 10^{-6}}{1} + \dfrac{-2 \times 10^{-6}}{2} + \dfrac{3 \times 10^{-6}}{3} \right)$
$V = 9 \times 10^9 \times 10^{-6} (1 - 1 + 1)$
$V = 9 \times 10^3\ \text{V}$
Answer: $9 \times 10^3\ \text{V}$
$V = \dfrac{1}{4\pi\epsilon_0} \left( \dfrac{q_1}{r_1} + \dfrac{q_2}{r_2} + \dfrac{q_3}{r_3} \right)$
The distances of the charges from the origin are:
$r_1 = \sqrt{1^2 + 0^2 + 0^2} = 1\ \text{m}$
$r_2 = \sqrt{0^2 + 2^2 + 0^2} = 2\ \text{m}$
$r_3 = \sqrt{0^2 + 0^2 + 3^2} = 3\ \text{m}$
Substituting the values of charges and distances:
$V = 9 \times 10^9 \left( \dfrac{1 \times 10^{-6}}{1} + \dfrac{-2 \times 10^{-6}}{2} + \dfrac{3 \times 10^{-6}}{3} \right)$
$V = 9 \times 10^9 \times 10^{-6} (1 - 1 + 1)$
$V = 9 \times 10^3\ \text{V}$
Answer: $9 \times 10^3\ \text{V}$
Hence, the correct answer is option: 'D'.