Question

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 \(\left\{\frac{1}{4\pi \epsilon _{0}}=9\times 10^{9}\ \text{SI unit}\right\}\)

• A. \(10^{6}\ \text{V}\)
• B. \(3\times 10^{3}\ \text{V}\)
• C. \(6\times 10^{6}\ \text{V}\)
• D. \(9\times 10^{3}\ \text{V}\)

Hint:

Potential is a scalar, so add kq/r for each charge using its distance from the origin.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The potential due to a point charge is \(V=\frac{1}{4\pi\epsilon_{0}}\frac{q}{r}\), and potentials add algebraically.

Step 2: Calculation
Distances from the origin: \(r_{1}=1\ \text{m}\), \(r_{2}=2\ \text{m}\), \(r_{3}=3\ \text{m}\). \(V=9\times 10^{9}\times 10^{-6}\left(\frac{1}{1}-\frac{2}{2}+\frac{3}{3}\right)\).

Step 3: Compare the options
\(V=9\times 10^{3}\times(1-1+1)=9\times 10^{3}\ \text{V}\), which is the fourth option.

Final Answer:
The potential at the origin is \(9\times 10^{3}\ \text{V}\). \[ \boxed{\text{(D) }9\times10^{3}\ \text{V}} \]