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?


Expert's answer

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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