The work done by an electric field when moving a charge on an equipotential surface is zero
because there is no change in electric potential energy. Electric lines of force are always
perpendicular to equipotential surfaces, meaning that any movement along the surface does
not alter the electric potential, confirming that no work is done.
The Correct answer is: Both (A) and (R) are correct and (R) is the correct explanation of (A)
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: