Question

Two metal sphere one of radius ' r ' and other of radius ' 2r ' both have same surface charge density ' G ' (sigma).They are brought in contact and seperated.What will be the new surface charge densities on them??

Answer 1

$$\begin{gathered} chargeonthefirstsphereisq=\sigma \times 4\pi r^2 \\ potentialonthefirstsphere{v}_1=\frac{kq}{r}=\frac{k\sigma \times 4\pi r^2}{r}=k\sigma \times 4\pi r \\ chargeonthesecondsphere=Q\hspace{0.17em}=\sigma \times 16\pi r^2 \\ {v}_2=\frac{kQ}{2r}=k\sigma \times 8\pi r \\ whentheyareincontact,chargewilltransferfrmhigherpotenmtialtolowerpotentialuntilpotentialofboththespheresbecomeequal. \\ {q}_{1}istheabountofchargetransfertherefore \\ \frac{k(q+q_1)}{r}=\frac{k(Q-q_1)}{2r} \\ 2(q+q_1)=(Q-q_1) \\ 3q_1=Q-2q \\ 3q_1=\sigma \times 16\pi r^2-2\sigma \times 4\pi r^2=\sigma \times 8\pi r^2 \\ {q}_1=\sigma \times 8\pi r^2/3 \\ {\sigma }_1\text{'}=\frac{q+q_1}{4\pi r^2}=\frac{\sigma \times 4\pi r^2+\sigma \times 8\pi r^2/3}{4\pi r^2}=\frac{5\sigma }{3} \\ {\sigma }_2\text{'}=\frac{Q-q_1}{16\pi r^2}=\frac{\sigma \times 16\pi r^2-\sigma \times 8\pi r^2/3}{16\pi r^2}=\frac{5\sigma }{6} \end{gathered}$$