Question

A positive charge +q= +8.0*10^-8 C with mass m=5.5*10^-10 kg, located at the +V plate, is at rest. If the final speed of the charge is vf=600m/s when it strikes the -Vplate, what is the electric potential V of the plates? A. 24.9 V B. 1240 V C. 1000 V D. 619 V E. 1.03 V

Answer 1

B. 1240 V

Step-by-step explanation:

Given

Initial velocity of the charge, u = 0 m/s

Final velocity of the charge, v = 600 m/s

Magnitude of the charge, q =
`8.0 * 10^(-8)C`

mass of the charge, m =
`5.5 * 10^(-10) kg`

Solution

Gain in mechanical energy = Loss in electric potential energy

`(1)/(2) mv^(2) - (1)/(2) mu^(2)= Vq\\\\0.5 * 5.5 * 10^(-10) * 600^(2)  - 0 = V * 8.0 * 10^(-8)\\\\V = 1237.5 V`

Rounding off the answer to closest tens, we get

V = 1240 V

Answer 2

`\Delta V = 1240 Volts`

Step-by-step explanation:

As we know that charge moves from higher potential to lower potential

so here change in electrostatic potential energy must be equal to final kinetic energy of the charge

so we will have

`q\Delta V = (1)/(2)mv^2`

so we will have

`q = 8* 10^(-8) C`

`m = 5.5 * 10^(-10) kg`

`v_f = 600 m/s`

`(8 * 10^(-8))\Delta V = (1)/(2)(5.5 * 10^(-10))(600^2)`

`\Delta V = 1240 Volts`