Question #106271

A cube of edge a carries a point charge q at each corner.

(a) Calculate resultant force on charge 3 by charges 5. 6 and e.

(b) Show that the resultant electric force on charge 2, by the rest of charges is given

Expert's answer

As per the given question,

The length of the side of the cube =a,

charges on each of the vertices = q

a)

distance between the vertices 3 and 8 =a2+a2=a2\sqrt{a^2+a^2}=a\sqrt{2}

distance between the vertices 3 and 6 =a2+a2=a2\sqrt{a^2+a^2}=a\sqrt{2}

distance between the vertices 3 and 5 = a2+(a2)2=a1+2=a3\sqrt{a^2+(a\sqrt{2})^2}=a\sqrt{1+2}=a\sqrt{3}

Force at the vertices 3, due to the vertices 5, 6 and 8

F=F35+F83+F63F=F_{35}+F_{83}+F_{63}


F=q24πϵo2a2(cos45i^sin45k^)+q24πϵo2a2(cos45i^+sin45j^)+q24πϵo3a2(cos45i^+sin45j^+cos45k^)\Rightarrow F=\dfrac{q^2}{4\pi \epsilon_o 2a^2}(\cos 45^\circ\hat{i}-\sin{45^\circ}\hat{k})+\dfrac{q^2}{4\pi \epsilon_o 2a^2}(\cos 45^\circ\hat{i}+\sin{45^\circ}\hat{j})+\dfrac{q^2}{4\pi \epsilon_o 3a^2}(\cos 45^\circ\hat{i}+\sin{45^\circ}\hat{j}+\cos 45\hat{k})

F=q24πϵoa2(12+132)i^+q24πϵoa2(12+132)j^+q24πϵoa2(12+132)k^\Rightarrow F=\dfrac{q^2}{4\pi \epsilon_o a^2}(\dfrac{1}{\sqrt{2}}+\dfrac{1}{3\sqrt{2}})\hat{i}+\dfrac{q^2}{4\pi \epsilon_o a^2}(\dfrac{1}{\sqrt{2}}+\dfrac{1}{3\sqrt{2}})\hat{j}+\dfrac{q^2}{4\pi \epsilon_o a^2}(\dfrac{1}{\sqrt{2}}+\dfrac{1}{3\sqrt{2}})\hat{k}

ii)

Fx=kq2a2+kq22a2cos45F_{x}= \dfrac{kq^2}{a^2}+\dfrac{kq^2}{2a^2}\cos 45

Fy=kq2a2+kq22a2sin45F_{y}= \dfrac{kq^2}{a^2}+\dfrac{kq^2}{2a^2}\sin 45

Fnet=Fx2+Fy2=keq2a2(2+12)=1.9q24πϵoa2F_{net}=\sqrt{F_x^2+F_y^2}=k_e\dfrac{q^2}{a^2}(\sqrt{2}+\dfrac{1}{2})=\dfrac{1.9q^2}{4\pi \epsilon_o a^2}

Ke=14πϵoK_e= \dfrac{1}{4\pi \epsilon_o}



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