Two identical capacitors are connected in series. The charge Q supplied by a source versus the potential difference V across the combination is plotted, yielding a straight line with slope S₀ . A dielectric material of dielectric constant K is then fully inserted into one of the capacitors. The new Q-V graph for the combination has a slope of 1.6 S₀ . The value of K is :

Step-by-step solution

The slope of a Q-V graph represents the equivalent capacitance, C_ eq = Q V . Initially, two identical capacitors of capacitance C are in series. The equivalent capacitance is: C_ eq1 = C 2 Thus, the initial slope is S₀ = C 2 . When a dielectric of constant K is inserted into one capacitor, its capacitance becomes KC . The new equivalent capacitance of the series combination is: C_ eq2 = C KC C + KC = KC K+1 Given that the new slope is 1.6 S₀ , we have: C_ eq2 = 1.6 S₀ KC K+1 = 1.6 ( C 2 ) = 0.8 C Dividing both sid