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Question

Let E$$_1$$(r), E$$_2$$(r) and E$$_3$$(r) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density $$\lambda$$, and an infinite plane with uniform surface charge density $$\sigma$$. If E$$_1(r_0) = E_2 (r_0) = E_3 (r_0)$$ at a given distance r$$_0$$, then :


A
Q=4σπr20
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B
r0=λ2πσ
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C
E1(r0/2)=2E2(r0/2)
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D
E2(r0/2)=4E3(r0/2)
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Solution

The correct option is C $$E_1 (r_0 / 2) = 2E_2 (r_0/2)$$
$$\displaystyle \frac{Q}{4 \pi \varepsilon_0r_0^2}=\frac{\lambda}{2 \pi \varepsilon_0 r_0} = \frac{\sigma}{2 \varepsilon_0}$$
$$\displaystyle E_1 \left ( \frac{r_0}{2} \right ) = \frac{Q}{\pi \varepsilon_0r_0^2}, E_2 \left ( \frac{r_0}{2} \right ), E_3 \left ( \frac{r_0}{2} \right ) = \frac{\sigma}{2 \varepsilon_0}$$
$$\displaystyle \therefore E_1 \left ( \frac{r_0}{2} \right ) = 2E_2 \left ( \frac{r_0}{2}\right)$$

Physics

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