A proton is released from rest in a uniform electric field which is directed along the x axis and of magnitude E=2*10exponent5 V/m.The proton undergoes a displacement from point A at the origin to point B at x=0.2m.What is the change in the potential energy of the proton between points A and B?

Remember that energy is conserved, so Ek + Ep will always equal Etotal.

What is your work so far? Let me know where you're stuck and I can help out.

hi Esskay,can u help me with the answer

To find the change in potential energy of the proton between points A and B, we need to calculate the work done on the proton by the electric field.

The potential energy (PE) of a charged particle in an electric field is given by the equation:

PE = q * ∆V

where q represents the charge of the particle in coulombs, and ∆V is the change in electric potential (voltage) between the two points.

In this case, the charge of the proton, q, is equal to the elementary charge, which is approximately 1.6 * 10^(-19) C.

To find the change in electric potential (∆V), we can use the equation:

∆V = E * ∆x

where E is the magnitude of the electric field in volts per meter (V/m), and ∆x is the displacement of the proton along the x-axis.

Plugging in the given values, we have:

E = 2 * 10^5 V/m
∆x = 0.2 m

So, ∆V = (2 * 10^5 V/m) * (0.2 m) = 4 * 10^4 V

Now, we can calculate the change in potential energy:

PE = q * ∆V = (1.6 * 10^(-19) C) * (4 * 10^4 V) = 6.4 * 10^(-15) J

Therefore, the change in potential energy of the proton between points A and B is 6.4 * 10^(-15) joules.