Question

Charges q are placed at the corners of a cube of side a. Find the potential at a) the centre b) centre of face c) mid point of an edge d) if q = 5 uc & a = 1 m , how much energy will be required to add 1 uc at one of the corners?

Answer 1

Dear student

(a) The distance of the centre to the all charges are same so the potential at the centre is given by
$$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_o}\frac{8q}{r} \\ butr=\frac{a\sqrt{3}}{2}so \\ V=\frac{1}{4\pi {\epsilon }_o}\frac{8q}{\frac{a\sqrt{3}}{2}} \\ V=\frac{4}{\sqrt{3}\pi {\epsilon }_o}\frac{q}{a} \end{gathered}$$
(b) At the centre of any face
$$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_o}\frac{4q}{r_1}+\frac{1}{4\pi {\epsilon }_o}\frac{4q}{r_2} \\ where{r}_{1}isthedis\tan ceofthechargeatthecornerofthefacetocentreofthatfaceand \\ {r}_{2}isthedis\tan ceofchargeatthecornerofotherfaceofcubetothecentreofface \\ here{r}_1=\frac{a}{\sqrt{2}}and{r}_2=\sqrt{\frac{3}{2}}a \\ soV=\frac{1}{4\pi {\epsilon }_o}\frac{4\sqrt{2}q}{a}+\frac{1}{4\pi {\epsilon }_o}4\sqrt{\frac{3}{2}}\frac{q}{a} \\ V=\frac{\sqrt{2}}{\pi {\epsilon }_o}\frac{q}{a}+\frac{1}{\pi {\epsilon }_o}\sqrt{\frac{3}{2}}\frac{q}{a} \\ V=\frac{1}{\pi {\epsilon }_o}\frac{q}{a}\left(\sqrt{2}+\sqrt{\frac{3}{2}}\right) \end{gathered}$$
(c) The potential at the midpoint of an edge
$$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_o}\frac{2q}{r_1}+\frac{1}{4\pi {\epsilon }_o}\frac{4q}{r_2}+\frac{1}{4\pi {\epsilon }_o}\frac{2q}{r_2} \\ here{r}_1=\frac{a}{2},r_2=\frac{\sqrt{5}}{2}aand{r}_3=\frac{3}{2}a \\ soV=\frac{1}{4\pi {\epsilon }_o}\frac{2\times 2q}{a}+\frac{1}{4\pi {\epsilon }_o}\frac{2\times 4q}{a\sqrt{5}}+\frac{1}{4\pi {\epsilon }_o}\frac{2\times 2q}{3a} \\ V=\frac{1}{\pi {\epsilon }_o}\frac{q}{a}\left(1+\frac{2}{\sqrt{5}}+\frac{1}{3}\right) \\ V=\frac{1}{\pi {\epsilon }_o}\frac{q}{a}\left(\frac{2}{\sqrt{5}}+\frac{4}{3}\right) \end{gathered}$$
(d)
$$\begin{gathered} Thsepotentialatonecornerofthecubewillbe \\ V=\frac{1}{4\pi {\epsilon }_o}\frac{3q}{r_1}+\frac{1}{4\pi {\epsilon }_o}\frac{3q}{r_2}+\frac{1}{4\pi {\epsilon }_o}\frac{q}{r_3} \\ here{r}_1=a,r_2=\sqrt{2}aand{r}_3=\sqrt{3}a \\ soV=\frac{1}{4\pi {\epsilon }_o}\frac{3q}{a}+\frac{1}{4\pi {\epsilon }_o}\frac{3q}{\sqrt{2}a}+\frac{1}{4\pi {\epsilon }_o}\frac{q}{\sqrt{3}a} \\ V=\frac{1}{4\pi {\epsilon }_o}\frac{q}{a}\left(3+\frac{3}{\sqrt{2}}+\frac{1}{\sqrt{3}}\right) \\ hereq=5\mu Canda=1mso \\ V=9\times {10}^9\times \frac{5\times {10}^{-6}}{1}\left(3+2.12+0.577\right) \\ V=9\times {10}^9\times 5\times {10}^{-6}\times 5.7 \\ V=2.56\times {10}^{5}V \\ Hencetheenergyrequiredtotakeachargeof1\mu Cfrominfinitytothiscorner \\ is \\ ∆U=q_o\left(V-V_{\infty }\right) \\ ∆U=1\times {10}^{-6}\times \left(2.56\times {10}^5-0\right) \\ ∆U=0.256J \end{gathered}$$