At a particular instant a proton is at the origin, moving with velocity < 4 104, 0, 0 > m/s. At this instant: (a) What is the electric field at location < -3 10-3, -3 10-3, 0 > m, due to the proton? (b) What is the magnetic field at the same location due to the proton? (c) How would the answers to (a) and (b) change if the particle had been an electron instead of a proton?

a) E = 7.57 10⁻⁵ N / C, b)  B= 2.11 10⁻¹⁸ T, c) the magnitudes remain the same, but the sign of the responses change

Explanation:

a) The electric field due to a charged particle is

            E = k q / r²

Let's look for the distance from the origin where the particle is to the point of interest

           r² = (-3 10⁻³ -0)² + (-3 10⁻³ -0)²

           r² = 19 10⁻⁶

The proton charge is

           q = +1.6 10⁻¹⁹ C

Let's calculate

           E = 8,988 10⁹  1.6 10⁻¹⁹ / 19 10⁻⁶

           E = 7.57 10⁻⁵ N / C

b) to calculate the magnetic field let's use the Biot-Savat law

           B = μ₀/4π     I / r²

The current is defined as the number of carriers per unit of time at a given point

           I = q v

Let's calculate

           B = 4π 10⁻⁷ /4π   (1.6 10⁻¹⁹  4 10⁴) / 19 10⁻⁶

            B= 2.11 10⁻¹⁸ T

c) an electron has a charge of the same magnitude, but in the opposite direction, negative.

   Therefore the magnitudes remain the same, but the senses (sign) of the responses change

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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i think the answer would be b