A body of mass 1 ~g and carrying a charge 10^ | vedclass

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(A) Given:Mass $m = 1 ~g = 10^{-3} ~kg$Charge $q = 10^{-8} ~C$Potential at $P$,$V_P = 600 ~V$Potential at $Q$,$V_Q = 0 ~V$Velocity at $Q$,$v_Q = 20 ~c

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$\sqrt{0.028}$

Vedclass · Answer

(A) Given:Mass $m = 1 ~g = 10^{-3} ~kg$Charge $q = 10^{-8} ~C$Potential at $P$,$V_P = 600 ~V$Potential at $Q$,$V_Q = 0 ~V$Velocity at $Q$,$v_Q = 20 ~cm/s = 0.2 ~m/s$According to the work-energy theorem,the work done by the electric field is equal to the change in kinetic energy:$W = \Delta KE = KE_Q - KE_P$$q(V_P - V_Q) = \frac{1}{2} m v_Q^2 - \frac{1}{2} m v_P^2$Substituting the values:$10^{-8} (600 - 0) = \frac{1}{2} (10^{-3}) (0.2)^2 - \frac{1}{2} (10^{-3}) v_P^2$$600 	imes 10^{-8} = 0.5 	imes 10^{-3} 	imes 0.04 - 0.5 	imes 10^{-3} v_P^2$$6 	imes 10^{-6} = 2 	imes 10^{-5} - 0.5 	imes 10^{-3} v_P^2$$0.5 	imes 10^{-3} v_P^2 = 20 	imes 10^{-6} - 6 	imes 10^{-6}$$0.5 	imes 10^{-3} v_P^2 = 14 	imes 10^{-6}$$v_P^2 = \frac{14 	imes 10^{-6}}{0.5 	imes 10^{-3}} = 28 	imes 10^{-3} = 0.028$$v_P = \sqrt{0.028} ~m/s$

$A$ body of mass $1 ~g$ and carrying a charge $10^{-8} ~C$ passes through two points $P$ and $Q$. $P$ and $Q$ are at electric potentials $600 ~V$ and $0 ~V$,respectively. The velocity of the body at $Q$ is $20 ~cm/s$. Its velocity in $m/s$ at $P$ is:

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