TS EAMCET · Physics · Electrostatics
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 \(\frac{q a}{4 \pi \varepsilon_0 z^2}\)
- B \(\frac{q}{4 \pi \varepsilon_0 a}\)
- C \(\frac{2 q a}{4 \pi \varepsilon_0\left(z^2-a^2\right)}\)
- D \(\frac{2 q a}{4 \pi \varepsilon_0\left(z^2+a^2\right)}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 q a}{4 \pi \varepsilon_0\left(z^2+a^2\right)}\)
Step-by-step Solution
Detailed explanation
Potential at \(P\) due to \((+q)\) charge \(V_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{(z-a)}\) Potential at \(P\) due to \((-q)\) charge \(V_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{-q}{(z+a)}\) Total potential at \(P\) due to \((A B)\) electric dipole…
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