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baherus [9]
3 years ago
7

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.
Physics
1 answer:
Assoli18 [71]3 years ago
4 0

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

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