a proton is moved so that its electric potential energy increases from 4.0×10⁻¹⁴ j to 9.0×10⁻¹⁴ j. the…

a proton is moved so that its electric potential energy increases from 4.0×10⁻¹⁴ j to 9.0×10⁻¹⁴ j. the magnitude of the charge on a proton is 1.602×10⁻¹⁹ c. what is the electric potential difference through which the proton moved? o 2.5×10⁵ v o 3.1×10⁵ v o 5.6×10⁵ v o 8.1×10⁵ v

a proton is moved so that its electric potential energy increases from 4.0×10⁻¹⁴ j to 9.0×10⁻¹⁴ j. the magnitude of the charge on a proton is 1.602×10⁻¹⁹ c. what is the electric potential difference through which the proton moved? o 2.5×10⁵ v o 3.1×10⁵ v o 5.6×10⁵ v o 8.1×10⁵ v

Answer

Explanation:

Step1: Calculate change in electric - potential energy

The change in electric - potential energy $\Delta U$ is given by $\Delta U=U_f - U_i$. Here, $U_i = 4.0\times10^{-14}\text{ J}$ and $U_f=9.0\times10^{-14}\text{ J}$. So, $\Delta U=(9.0\times 10^{-14}-4.0\times 10^{-14})\text{ J}=5.0\times 10^{-14}\text{ J}$.

Step2: Use the formula for electric - potential difference

The relationship between the change in electric - potential energy $\Delta U$, the charge $q$, and the electric - potential difference $\Delta V$ is $\Delta U = q\Delta V$. We know $q = 1.602\times10^{-19}\text{ C}$ and $\Delta U=5.0\times 10^{-14}\text{ J}$. Solving for $\Delta V$, we get $\Delta V=\frac{\Delta U}{q}$. Substitute the values: $\Delta V=\frac{5.0\times 10^{-14}\text{ J}}{1.602\times 10^{-19}\text{ C}}\approx3.1\times 10^{5}\text{ V}$.

Answer:

$3.1\times 10^{5}\text{ V}$