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QJEEM24Ph7014: JEE Main 2024 Physics - Class 12 JEEM24Ph7014 Solution: The distance between charges and is and between and is . The…

The distance between charges +q+q and −q-q is 2l2 l and between +2q+2 q and −2q-2 q is 4l4 l . The electrostatic potential at point PP at a distance rr from center OO is −α[qlr2]×109 V-\alpha\left[\frac{q l}{r^2}\right] \times 10^9 \mathrm{~V} , where the value of α\alpha is __________. (Use 14πε0=9×109 Nm2C−2\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~Nm}^2 \mathrm{C}^{-2} )

JEE Main 2024 (Online) 31st January Evening Shift Physics - Electrostatics Question 42 English

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JEE Main 2024 (Online) 31st January Evening Shift Physics - Electrostatics Question 42 English Explanation

amp;V=Kp→⋅r→r3=9×109(6qℓ)r2cos⁡(120∘)amp;=−(27)(qℓr2)×109Nm2c−2amp;⇒α=27\begin{aligned}& \mathrm{V}=\frac{\mathrm{K} \overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}}{\mathrm{r}^3}=\frac{9 \times 10^9(6 \mathrm{q} \ell)}{\mathrm{r}^2} \cos \left(120^{\circ}\right) \\\\& =-(27)\left(\frac{\mathrm{q} \ell}{\mathrm{r}^2}\right) \times 10^9 \mathrm{Nm}^2 \mathrm{c}^{-2} \\\\& \Rightarrow \alpha=27\end{aligned}

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