I need help figuring out this practice problem. Please provide a complete detail

Question

I need help figuring out this practice problem. Please provide a complete detailed explanation. Thanks.
An electron (q = -e, m = 9.1e-31 kg) is projected out along the + x axis with an initial speed of (m*3)x10^6 m/s. It goes 45 cm and stops due to a uniform electric field in the region. Find the magnitude and direction of the field.
ans: +57 N/C
I need help figuring out this practice problem. Please provide a complete detailed explanation. Thanks.
An electron (q = -e, m = 9.1e-31 kg) is projected out along the + x axis with an initial speed of (m*3)x10^6 m/s. It goes 45 cm and stops due to a uniform electric field in the region. Find the magnitude and direction of the field.
ans: +57 N/C

Explanation / Answer

Given The initial velocity of the electron u =3*10^6 m/s        Final velocity of the electron v = 0.0 m/s the distance travelled by the electron s =45*10^-2 m ------------------------------------------------------- Using kinematic relation v^2 -u^2 =2as                                     (0.0 m/s)^2 -(3*10^6 m/s)^2 = 2*a* 45*10^-2 m The acceleration of the electron a = 10^13 m/s^2 According to Newton's second law of motion F = m*a The force acting on the electron F = (9.1*10^-31 kg)*(10^13) N F = 9.1*10^-18 N the relation between force and electric field is F = q*E Here q is charge of the electron =1.6*10^-19 C The magnitude of the electric field E =F/q =9.1*10^-18/9*1.6*10^-19 E = 56.8 N/C ----------------------------------------------------------- Always electric filed directs from positive to negative region Since electron has negative charge it accelerates opposite to the elecrtic field Hence the field direction is along negative x-axis

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