Question

Two point charges \(-q\) and \(+q\) are located at points \((0,0,-a)\) and \((0,0,a)\), respectively. The electric potential at a point \((0,0,z)\), where \(z>a\) is

• A. \(\dfrac{qa}{4\pi\varepsilon_0 z^2}\)
• B. \(\dfrac{q}{4\pi\varepsilon_0 a}\)
• C. \(\dfrac{2qa}{4\pi\varepsilon_0 (z^2-a^2)}\)
• D. \(\dfrac{2qa}{4\pi\varepsilon_0 (z^2+a^2)}\)

Hint:

Potential is scalar, so add algebraically. For charges on z-axis: use \(r=z\pm a\) and simplify fractions.

Answer 1

The Correct Option is C

Step 1: Write potential due to each charge.
Potential due to point charge:
\[ V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} \]
Step 2: Compute distances from point \((0,0,z)\).
Distance from \(+q\) at \((0,0,a)\):
\[ r_+ = z-a \]
Distance from \(-q\) at \((0,0,-a)\):
\[ r_- = z+a \]
Step 3: Total potential.
\[ V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{z-a} + \frac{-q}{z+a}\right) \]
Step 4: Simplify.
\[ V = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{z-a} - \frac{1}{z+a}\right) \]
\[ V = \frac{q}{4\pi\varepsilon_0}\left(\frac{(z+a)-(z-a)}{(z-a)(z+a)}\right) \]
\[ V = \frac{q}{4\pi\varepsilon_0}\left(\frac{2a}{z^2-a^2}\right) \]
\[ V = \frac{2qa}{4\pi\varepsilon_0 (z^2-a^2)} \]
Final Answer: \[ \boxed{\dfrac{2qa}{4\pi\varepsilon_0 (z^2-a^2)}} \]