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 ="\\sqrt{a^2+a^2}=a\\sqrt{2}"
distance between the vertices 3 and 6 ="\\sqrt{a^2+a^2}=a\\sqrt{2}"
distance between the vertices 3 and 5 = "\\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=F_{35}+F_{83}+F_{63}"
"\\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})"
"\\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)
"F_{x}= \\dfrac{kq^2}{a^2}+\\dfrac{kq^2}{2a^2}\\cos 45"
"F_{y}= \\dfrac{kq^2}{a^2}+\\dfrac{kq^2}{2a^2}\\sin 45"
"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}"
"K_e= \\dfrac{1}{4\\pi \\epsilon_o}"
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