A uniform electric field of magnitude 375 N/C pointing in the positive x-direction acts on a negative charge, which is initially at rest. After the electron has moved 3.20 cm, calculate, (a) The work done by the field on the electron. (b) The change in po

(a) The work done by the electric field on the electron is given by the formula: {eq}W = vec{F}cdot vec{d} {/eq} where {eq}vec{F}{/eq} is the force exerted by the electric field on the electron, and {eq}vec{d}{/eq} is the displacement of the electron. Since the electric field is uniform, the force on the electron is given by: {eq}vec{F} = qvec{E} {/eq} where {eq}q{/eq} is the charge of the electron and {eq}vec{E}{/eq} is the electric field. Since the electron is negative, its charge is {eq}-1.6times 10^{-19} \text{ C}{/eq}, and the force on it is: {eq}vec{F} = (-1.6times10^{-19}\text{ C})(375\text{ N/C})hat{imath} = -6times10^{-17}\text{ N}hat{imath} {/eq} where {eq}hat{imath}{/eq} is a unit vector in the positive {eq}x{/eq}-direction. The displacement of the electron is {eq}vec{d} = (3.20\text{ cm})hat{imath} = 0.032\text{ m}hat{imath}{/eq}. Therefore, the work done by the electric field on the electron is: {eq}W = vec{F}cdotvec{d} = (-6times10^{-17}\text{ N}hat{imath})cdot(0.032\text{ m}hat{imath}) = -1.92times10^{-18}\text{ J} {/eq} The negative sign in the answer indicates that the work done by the electric field is negative, which means that the electric field has done negative work on the electron. This is because the direction of the force on the electron is opposite to the direction of the displacement of the electron. (b) The change in potential energy associated with the electron is given by the formula: {eq}Delta U = -W {/eq} where {eq}W{/eq} is the work done by the electric field. Therefore, the change in potential energy is: {eq}Delta U = -(-1.92times10^{-18}\text{ J}) = 1.92times10^{-18}\text{ J} {/eq} The positive sign in the answer indicates that the potential energy of the electron has increased, which means that the electron has moved to a region of higher potential. (c) Since the electron starts from rest, its initial kinetic energy is zero. According to the work-energy theorem, the work done on the electron by the electric field must be equal to the change in kinetic energy of the electron: {eq}W = Delta K {/eq} where {eq}Delta K{/eq} is the change in kinetic energy of the electron. Therefore, we have: {eq}frac{1}{2}mv^2 = -W = 1.92times10^{-18}\text{ J} {/eq} where {eq}v{/eq} is the velocity of the electron and {eq}m{/eq} is the mass of the electron. The mass of the electron is {eq}9.11times10^{-31}\text{ kg}{/eq}. Solving for {eq}v{/eq}, we get: {eq}v = sqrt{frac{2Delta K}{m}} = sqrt{frac{2(1.92times10^{-18}\text{ J})}{9.11times10^{-31}\text{ kg}}} \approx 8.95 times 10^5 \text{ m/s} {/eq} Therefore, the velocity of the electron after moving a distance of 3.20 cm under the influence of the uniform electric field of magnitude 375 N/C is \approx imately 8.95 ? 105 m/s in the positive x-direction.