The amount of work done in increasing the voltage across the plates of the capacitor from 5V to 10V is ’W’. The work done in increasing it from 10V to 15V will be
(A) W
(B) 0.6 W
(C) 1.25 W
(D) 1.67 W

Hint: Calculate the work done in increasing the voltage from 5V to 10V and from 10V to 15V separately now divide both the work to find the ratio of new work done W1in terms of W.

Complete step-by-step solution:
When a capacitor is charged by a voltage source like a battery it stores the electric energy. Since the work done is stored in the form of energy we can use its expression.
The energy stored in capacitor is given by
$E = \dfrac{1}{2}C{V^2}$
Work done during the change of potential from V1 to V2 is given by
\[W = \dfrac{1}{2}C\left[ {V_1^2 - V_2^2} \right]\]
Here,
V1 and V2 are the initial and final voltage respectively.
C is the capacitance
Work done during the change of voltage from 5V to 10V is
$
  W = \dfrac{1}{2}C\left[ {{{10}^2} - {5^2}} \right] \\
  W = \dfrac{1}{2}C\left[ {100 - 25} \right] \\
  W = \dfrac{{75}}{2}C \\
 $ → (1)
Work done during the change of voltage from 10V to 15V is
$
  {W_1} = \dfrac{1}{2}C\left[ {{{15}^2} - {{10}^2}} \right] \\
  {W_1} = \dfrac{1}{2}C\left[ {225 - 100} \right] \\
  {W_1} = \dfrac{{125}}{2}C \\
 $ → (2)
Now dividing equation (2) by (1) we get,
$
  {W_1} = \dfrac{{125}}{{75}}W \\
  {W_1} = 1.67W \\
 $

Therefore, the work done in increasing from 10V to 15V is 1.67W and the correct option is D.

Note: The energy stored or the work done is also given by the formula
$
  E = \dfrac{1}{2}QV \\
  E = \dfrac{{{Q^2}}}{{2C}} \\
 $