A proton has an initial speed of 7·105 m/s. It is brought to rest by an electric field.

a.) What is the magnitude of the voltage difference needed to bring this charge to rest?

answer: 365.313*7v, WHY

i forgot to square the whole set of numbers. u=.5mv^2 and then u=qV

To find the magnitude of the voltage difference needed to bring the proton to rest, we can use the equation that relates the change in electrical potential energy (∆PE) to the charge (q) and the voltage difference (ΔV):

∆PE = q * ΔV

Since the proton comes to rest, its initial kinetic energy (∆KE) is equal to zero. The change in kinetic energy (∆KE) can be calculated using the formula:

∆KE = (1/2) * m * v^2

where m is the mass of the proton (approximately 1.67 x 10^(-27) kg) and v is the initial speed of the proton (7 x 10^5 m/s).

Now, ∆KE and ∆PE are related by the conservation of energy principle:

∆KE = - ∆PE

Since ∆KE is zero, we can rewrite the equation as:

0 = - q * ΔV

Solving for ∆V, we get:

ΔV = 0 / - q

The negative sign indicates that the electric field is acting in the opposite direction of the proton's motion to bring it to rest. Now substituting the charge of the proton (approximately 1.6 x 10^(-19) C) into the equation, we have:

ΔV = 0 / (-1.6 x 10^(-19) C) = 0 V

This means that the magnitude of the voltage difference needed to bring the proton to rest is 0 Volt. However, it is important to note that this result could be due to rounding errors or approximations in the given values. It is always recommended to consult the original question or additional information to clarify any uncertainties.