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

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(C) Using the work-energy theorem,the change in kinetic energy is equal to the work done by the electric field:$\Delta K = W_{elec}$$\frac{1}{2}m(v_B^

Vedclass · Accepted Answer

$16.8$

Vedclass · Answer

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

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

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