Question

a proton travelling along is x-axis is slowed by a uniform electric field E. at x = 20.0 cm, the proton has a speed of 3.5x10^6 m/s and at 80.0 cm the speed is zero. Determine the magnitude and direction of e,

Answer 1

`E=101955.8volt/m`

The electric field direction is toward the right

Step-by-step explanation:

From the question we are told that

Initial X-co-ordinate of proton
`X_1=20.0cm => (20)/(100)m`

Initial speed of Proton
`V_1= 3.5*10^6 m/s`

Final X-co-ordinate of proton
`X_2=80.0cm => (80.0)/(100)m`

Final speed of Proton
`V_2=0`

Generally the mass of Proton is given by

`m_P=1.67*10^-27`

Generally the kinetic energy of the proton is mathematically given by

`K.E_p=1/2mv^2`

`K.E_p=1/2*1.6*10^-^2^7*(3.5*10^6)^2`

`K.E_p=9.8*10^-^1^5`

Generally the change in electric potential
`\triangle V` is mathematically given by

`\triangle V =(K.E_p)/(q)`

Charge on a proton
`q=1.602*10^-^1^9`

`\triangle V =(9.8*10^-^1^5)/(1.602*10^-^1^9)`

`\triangle V =61173.5volts`

Generally the equation for magnitude of an electric field is mathematically given by

`E=(\triangle V)/(\triangle d)`

Where

`d=0.8m-0.2m\\d=0.6m`

Therefore

`E=(61173.5)/(0.8-0.2)`

`E=(61173.5)/(0.6)`

`E=101955.8volt/m`

The direction of the charge is towards the right