Question

A particle with a charge of -4.0 μC and a mass of 3.2 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 72 m/s. The only force acting on the particle is the electric force. What is the potential difference VB - VA between A and B? If VB is greater than VA, then give the answer as a positive number. If VB is less than VA, then give the answer as a negative number.

Answer 1

Answer:

$V_B-V_A=-20736-0=-20736volt$

Explanation:

We have given charge on the particle $q=-4\mu C=-4\times 10^{-6}C$

Mass of the charge particle $m=3.2\times 10^{-6}kg$

From energy of conservation kinetic energy will be equal to potential energy

So at point A

$\frac{1}{2}mv^2=qV$

At point a velocity is zero

So $\frac{1}{2}(3.2\times10^{-6} )0^2=-4\times 10^{-6}V_a$

$V_A=0volt$

At point B velocity will be 72 m/sec

So $\frac{1}{2}\times 3.2\times 10^{-6}72^2=-4\times 10^{-6}V_b$

$V_B=-20736volt$

So $V_B-V_A=-20736-0=-20736volt$