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An electron is near a proton exhibiting a non-uniform electric field. The electron moves to a distance of 2. 5 nm from the proton. What is the electrical potential energy of the electron in this point charge? (-9. 21 x 10-20 J)

(have answer, need work)

User Yohannes
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1 Answer

2 votes

Answer:

Approximately
9.2 * 10^(-20)\; {\rm J} assuming that both particles behave like point charges.

Step-by-step explanation:

In a system consisting of two point charges
q_(1) and
q_(2) at a distance of
r apart from each other, the electrical potential energy of the system would be:


\displaystyle (k\, q_(1)\, q_(2))/(r),

Where
k \approx 8.99* 10^(9)\; {\rm N\cdot m^(2) \cdot C^(-2)} is the Coulomb Constant.

The electric charge on each proton is equal to the elementary charge: approximately
1.602 * 10^(-19)\; {\rm C}. Electric charge on each electron has the same magnitude but the opposite sign: approximately
(-1.602) * 10^(-19)\; {\rm C}

Therefore, in this question, the two charges in the system are:


  • q_(1) \approx 8.99 * 10^(9)\; {\rm N\cdot m^(2) \cdot C^(-2)} for the proton, and

  • q_(2) \approx (-8.99) * 10^(9)\; {\rm N\cdot m^(2) \cdot C^(-2)}.

Apply unit conversion and ensure that the distance between the two particles is in the standard unit of meters:


\begin{aligned} r &= 2.5\; {\rm nm} * \frac{1\; {\rm m}}{10^(9)\; {\rm nm}} = 2.5 * 10^(-9)\; {\rm m} \end{aligned}.

Substitute these values into the equation for the electric potential:


\begin{aligned} & (k\, q_(1)\, q_(2))/(r) \\ \approx \; & ((1.602 * 10^(-19))\, (8.99 * 10^(9))\, (-8.99 * 10^(9)))/((2.5 * 10^(-9)))\; {\rm J} \\ \approx\; & (-9.2) * 10^(-20)\; {\rm J}\end{aligned}.

In other words, the electric potential of this system would be approximately
9.2 * 10^(-20)\; {\rm J}.

User Aero Chocolate
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