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 =a 2 + a 2 = a 2 \sqrt{a^2+a^2}=a\sqrt{2} a 2 + a 2 = a 2
distance between the vertices 3 and 6 =a 2 + a 2 = a 2 \sqrt{a^2+a^2}=a\sqrt{2} a 2 + a 2 = a 2
distance between the vertices 3 and 5 = a 2 + ( a 2 ) 2 = a 1 + 2 = a 3 \sqrt{a^2+(a\sqrt{2})^2}=a\sqrt{1+2}=a\sqrt{3} a 2 + ( a 2 ) 2 = a 1 + 2 = a 3
Force at the vertices 3, due to the vertices 5, 6 and 8
F = F 35 + F 83 + F 63 F=F_{35}+F_{83}+F_{63} F = F 35 + F 83 + F 63
⇒ F = q 2 4 π ϵ o 2 a 2 ( cos 4 5 ∘ i ^ − sin 4 5 ∘ k ^ ) + q 2 4 π ϵ o 2 a 2 ( cos 4 5 ∘ i ^ + sin 4 5 ∘ j ^ ) + q 2 4 π ϵ o 3 a 2 ( cos 4 5 ∘ i ^ + sin 4 5 ∘ j ^ + cos 45 k ^ ) \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 = 4 π ϵ o 2 a 2 q 2 ( cos 4 5 ∘ i ^ − sin 4 5 ∘ k ^ ) + 4 π ϵ o 2 a 2 q 2 ( cos 4 5 ∘ i ^ + sin 4 5 ∘ j ^ ) + 4 π ϵ o 3 a 2 q 2 ( cos 4 5 ∘ i ^ + sin 4 5 ∘ j ^ + cos 45 k ^ )
⇒ F = q 2 4 π ϵ o a 2 ( 1 2 + 1 3 2 ) i ^ + q 2 4 π ϵ o a 2 ( 1 2 + 1 3 2 ) j ^ + q 2 4 π ϵ o a 2 ( 1 2 + 1 3 2 ) 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} ⇒ F = 4 π ϵ o a 2 q 2 ( 2 1 + 3 2 1 ) i ^ + 4 π ϵ o a 2 q 2 ( 2 1 + 3 2 1 ) j ^ + 4 π ϵ o a 2 q 2 ( 2 1 + 3 2 1 ) k ^
ii)
F x = k q 2 a 2 + k q 2 2 a 2 cos 45 F_{x}= \dfrac{kq^2}{a^2}+\dfrac{kq^2}{2a^2}\cos 45 F x = a 2 k q 2 + 2 a 2 k q 2 cos 45
F y = k q 2 a 2 + k q 2 2 a 2 sin 45 F_{y}= \dfrac{kq^2}{a^2}+\dfrac{kq^2}{2a^2}\sin 45 F y = a 2 k q 2 + 2 a 2 k q 2 sin 45
F n e t = F x 2 + F y 2 = k e q 2 a 2 ( 2 + 1 2 ) = 1.9 q 2 4 π ϵ o a 2 F_{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} F n e t = F x 2 + F y 2 = k e a 2 q 2 ( 2 + 2 1 ) = 4 π ϵ o a 2 1.9 q 2
K e = 1 4 π ϵ o K_e= \dfrac{1}{4\pi \epsilon_o} K e = 4 π ϵ o 1