C}_{1}}}{{{N}_{1}}}{{(4V)}^{2}}\] Case II. When the capacitors are joined in parallel \[{{U}_{parallel}}=\frac{1}{2}({{n}_{2}}{{C}_{2}}){{V}^{2}}\] Given, \[{{U}_{series}}={{U}_{parallel}}\] or \[\frac{1}{2}\frac{{{C}_{1}}}{{{n}_{1}}}{{(4V)}^{2}}=\frac{1}{2}({{n}_{2}}{{C}_{2}}){{V}^{2}}\] \[\Rightarrow \] \[{{C}^{2}}=\frac{16{{C}_{1}}}{{{n}_{2}}\,{{n}_{1}}}\]