Question #259369

An object having a mass of 10 g and a charge of + 8.0x10^-5 C is placed in an electric field defined by Ex = 3.0x10^3 N/C, Ey = - 6000 N/C, and Ez = 0. (a) What are the magnitude and direction of the force on the object? (b) If the object starts from rest at the origin, what will be its coordinates after 3.0 s?


1
Expert's answer
2021-11-01T17:14:46-0400

Mass(m)=10g

Charge (wa)=8.0×105C8.0\times10^{-5}C

Electric field


Ex=3×103N/C,Ey=6000N/C,Ez=0N/CE_x=3\times10^{3}N/C,E_y=-{6000}N/C,E_z=0N/C

E=Exi^+Eyj^+Ezk^E=E_x\hat{i}+E_y\hat{j}+E_z\hat{k}

E=3000i^6000j^+0k^E=3000\hat{i}-6000\hat{j}+0\hat{k}

Force

F=qEF=qE


F=8.0×105×(3000i^6000j^+0k^)F=8.0\times10^{-5}\times(3000\hat{i}-6000\hat{j}+0\hat{k})

F=(0.24i^0.48j^)NF=(0.24\hat{i}-0.48\hat{j} )N


F=(0.242++(0.48)2=0.288=0.5366N|F|=\sqrt{(0.24^2++(-0.48)^2}=\sqrt{0.288}=0.5366N

Direction


θ=tan1(0.480.24)=tan1(2)=63.43°\theta=tan^{-1}(\frac{-0.48}{0.24})=tan^{-1}({-2})=-63.43°

Part(b)

Force (F)=ma

a=Fma=\frac{F}{m}

a=0.5366102=53.66m/seca=\frac{0.5366}{10^{-2}}=53.66m/sec

We know that

S=ut+12at2S=ut+\frac{1}{2}at^2


X=0+12×53.66×32=241.47mX=0+\frac{1}{2}\times53.66\times3^2=241.47 m


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