A uniform electric field of 10,N/C is created | vedclass

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(D) Given:Electric field $E = 10\,N/C$Kinetic energy $K = 0.5\,eV = 0.5 	imes 1.6 	imes 10^{-19}\,J$Length of plates $L = 10\,cm = 0.1\,m$Initial ki

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$45$

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(D) Given:Electric field $E = 10\,N/C$Kinetic energy $K = 0.5\,eV = 0.5 	imes 1.6 	imes 10^{-19}\,J$Length of plates $L = 10\,cm = 0.1\,m$Initial kinetic energy $K = \frac{1}{2}mv_x^2 = 0.5\,eV$$v_x = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 	imes 0.5 	imes e}{m}} = \sqrt{\frac{e}{m}}$Time taken to cross the field: $t = \frac{L}{v_x}$Vertical velocity acquired: $v_y = a_y t = \left(\frac{eE}{m}ight) \left(\frac{L}{v_x}ight)$The angle of deviation $	heta$ is given by:$	an 	heta = \frac{v_y}{v_x} = \frac{eE L}{m v_x^2}$Since $K = \frac{1}{2}mv_x^2$,we have $mv_x^2 = 2K = 2(0.5\,eV) = 1\,eV = e\,J$$	an 	heta = \frac{eEL}{e} = EL$$	an 	heta = 10\,N/C 	imes 0.1\,m = 1$$	heta = 	an^{-1}(1) = 45^{\circ}$

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