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
1
Expert's answer
2020-03-24T09:57:51-0400

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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