TS EAMCET · Physics · Electrostatics
Along the X-axis, three charges \(\frac{q}{2},-q\) and \(\frac{q}{2}\) are placed at \(x=0, x=a\) and \(x=2 a\) respectively. The resultant electric potential at \(x=a+r\) (if \(a< < r)\) is : ( \(\varepsilon_0\) is the permittivity of free space)
- A \(\frac{q a}{4 \pi \varepsilon_0 r^2}\)
- B \(\frac{q a^2}{4 \pi \varepsilon_0 r^3}\)
- C \(\frac{q\left(\frac{a^2}{4}\right)}{4 \pi \varepsilon_0 r^3}\)
- D \(\frac{q}{4 \pi \varepsilon_0 r}\)
Answer & Solution
Correct Answer
(B) \(\frac{q a^2}{4 \pi \varepsilon_0 r^3}\)
Step-by-step Solution
Detailed explanation
We have to find the electric potential at point \(P\). \(V_P=\left[\frac{1}{4 \pi \varepsilon_0} \frac{q / 2}{(r+a)}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q / 2}{(r-a)}\right]\)…
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