A ball of mass 1, g and charge 10^{-8}, C mov | vedclass

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Question

(C) According to the work-energy theorem,the change in kinetic energy is equal to the work done by the electric field.$W = q(V_A - V_B) = \Delta K$$q(

Vedclass · Accepted Answer

$16.8\, cm/s$

Vedclass · Answer

(C) According to the work-energy theorem,the change in kinetic energy is equal to the work done by the electric field.$W = q(V_A - V_B) = \Delta K$$q(V_A - V_B) = \frac{1}{2}m(v_B^2 - v_A^2)$Given: $m = 1\, g = 10^{-3}\, kg$,$q = 10^{-8}\, C$,$V_A = 600\, V$,$V_B = 0\, V$,$v_B = 20\, cm/s = 0.2\, m/s$.Substituting the values:$10^{-8}(600 - 0) = \frac{1}{2} 	imes 10^{-3} 	imes ((0.2)^2 - v_A^2)$$6 	imes 10^{-6} = 0.5 	imes 10^{-3} 	imes (0.04 - v_A^2)$$12 	imes 10^{-3} = 0.04 - v_A^2$$v_A^2 = 0.04 - 0.012 = 0.028$$v_A = \sqrt{0.028} \approx 0.1673\, m/s \approx 16.73\, cm/s \approx 16.8\, cm/s$.

$A$ ball of mass $1\, g$ and charge $10^{-8}\, C$ moves from a point $A$,where the potential is $600\, V$,to a point $B$,where the potential is $0\, V$. The velocity of the ball at point $B$ is $20\, cm/s$. The velocity of the ball at point $A$ will be:

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