Two point charge -q and +q/2 are situated at origin and at point (a, 0, 0) respectively. point along

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1. Electric Charges and Fields

medium

Two point charge $-q$ and $+q/2$ are situated at the origin and at the point $(a, 0, 0)$ respectively. The point along the $X$ - axis where the electric field vanishes is

$x = \frac{a}{{\sqrt 2 }}$

$x = \sqrt 2 a$

$x = \frac{{\sqrt 2 a}}{{\sqrt 2 - 1}}$

$x = \frac{{\sqrt 2 a}}{{\sqrt 2 + 1}}$

Solution

$\frac{{kq}}{{{x^2}}} = \frac{{kq/2}}{{{{(x – a)}^2}}}$ or $2{(x – a)^2} = {x^2}$ or $\sqrt 2 (x – a) = x$
or $(\sqrt 2 – 1)x = \sqrt 2 \,a$ or $x = \left( {\frac{{\sqrt 2 \,a}}{{\sqrt 2 – 1}}} \right)$

Standard 12

Physics

The bob of a simple pendulum has mass $2\,g$ and a charge of $5.0\,\mu C$. It is at rest in a uniform horizontal electric field of intensity $2000\,\frac{V}{m}$. At equilibrium, the angle that the pendulum makes with the vertical is (take $g = 10\,\frac{m}{{{s^2}}}$)

hard

(JEE MAIN-2019)

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hard

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medium

Two point charge $-q$ and $+q/2$ are situated at the origin and at the point $(a, 0, 0)$ respectively. The point along the $X$ - axis where the electric field vanishes is

Solution

Similar Questions

The bob of a simple pendulum has mass $2\,g$ and a charge of $5.0\,\mu C$. It is at rest in a uniform horizontal electric field of intensity $2000\,\frac{V}{m}$. At equilibrium, the angle that the pendulum makes with the vertical is (take $g = 10\,\frac{m}{{{s^2}}}$)

The magnitude of electric field intensity $E$ is such that, an electron placed in it would experience an electrical force equal to its weight is given by

Two charges each equal to $\eta q({\eta ^{ – 1}} < \sqrt 3 )$ are placed at the corners of an equilateral triangle of side $a$. The electric field at the third corner is ${E_3}$ where $({E_0} = q/4\pi {\varepsilon _0}{a^2})$

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