The electric field in a cathode ray tube is supposed to accelerate electrons from 0 to 1.60 * 10⁷m/s in a distance of 4.00 cm. What electric field is required? (m_e = 9.11 * 10 ^- ³¹kg and e = 1.60 *10 ^- ¹⁹ C)

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The correct answer is A.Explanation:The electric field is given by:$$E = \frac{ma}{q}$$where:* m is the mass of the electron* a is the acceleration of the electron* q is the charge of the electronSubstituting the given values into the equation, we get:$$E = \frac{(9.11 	imes 10^{-31} 	ext{ kg})(1.60 	imes 10^7 	ext{ m/s}^2)}{1.60 	imes 10^{-19} 	ext{ C}} = 18,200 	ext{ N/C}$$

The Electric Field in a Cathode Ray Tube Is Supposed