A flat circular disc has a charge +Q uniforml | vedclass

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(D) The potential $V$ at a distance $x$ from the center of a uniformly charged disc of radius $R$ is given by $V(x) = \frac{\sigma}{2\epsilon_0} (\sqr

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Any of the above three situations is possible depending on the magnitude of $E$

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(D) The potential $V$ at a distance $x$ from the center of a uniformly charged disc of radius $R$ is given by $V(x) = \frac{\sigma}{2\epsilon_0} (\sqrt{x^2 + R^2} - x)$, where $\sigma$ is the surface charge density.As the charge $+q$ approaches the disc, it experiences a repulsive electrostatic force.The potential energy of the charge at distance $x$ is $U(x) = qV(x)$.By the law of conservation of energy, $E + U(\infty) = K(x) + U(x)$. Since $U(\infty) = 0$, we have $E = K(x) + qV(x)$.At the point of closest approach, $K(x) = 0$, so $E = qV(x)$.$1$. If $E$ is very large, the charge will hit the disc $(x=0)$.$2$. If $E$ is exactly equal to the potential energy at the surface, it will touch the disc and return.$3$. If $E$ is small, the charge will stop at some distance $x > 0$ and return back along its path without touching the disc.Thus, all three situations are possible depending on the magnitude of $E$.

$A$ flat circular disc has a charge $+Q$ uniformly distributed on it. $A$ charge $+q$ is thrown with kinetic energy $E$ towards the disc along its normal axis. The charge $q$ will:

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