Question

A particle with a charge of −1.7 μC and a mass of 3.0 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 43 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

`V_B-V_A=+1631.47 V`

Step-by-step explanation:

We are given that

`q=-1.7\mu C=-1.7* 10^(-6) C`

Mass,m=
`3.0* 10^(-6)kg`

Speed,v=43 m/s

We have to find the potential difference between A and B.

By law of conservation of energy

Kinetic energy=Potential energy

`(1)/(2)mv^2=q(V_B-V_A)`

Substitute the values

`(1)/(2)(3* 10^(-6)(43)^2=1.7* 10^(-6)(V_B-V_A)`

`V_B-V_A=(3* 10^(-6))/(2* 1.7* 10^(-6))(43)^2`

`V_B-V_A=1631.47 V`

Negative charge is always accelerated towards high potential.Therefore,potential at point B is higher than the potential at point A.

Hence,
`V_B-V_A=+1631.47 V`