The drawing shows a loudspeaker a and point c, where a listener is positioned. a second loudspeaker b is located somewhere to the right of loudspeaker a. both speakers vibrate in phase and are playing a 67.6-hz tone. the speed of sound is 343 m/s. what is the closest to speaker a that speaker b can be located, so that the listener hears no sound? (assume |ac| = 4.00 m.)

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.