Two protons arc held at a separation of 2.000 times 10^-10 m. How much electric

Question

Two protons arc held at a separation of 2.000 times 10^-10 m. How much electric potential energy do they have? (Use k = 9.00 times 10^9 N middot m^2/C^2) 5.760 times 10-9 J 1.152 times 10^-18 J 3.467 times l0^-18 J 3.467 times 10^-18 J 7.200 times 10^10 J The same two protons of question 13 arc now pushed until their separation is 3.000 x 10*12 m. What was the change in potential energy of the protons? (Use it = 9.00 times 10^9 N middot m^2/C^2.) 7.565 times 10^-17 J, an increase 7.565 times 10^-17 J, a decrease 7.680 times 10^-17, an increase 7.680 times 10-17, a decrease Questions 15 to 18. Consider the two point charges illustrated here, held 0.0393 m apart firm each other. How much electric potential energy do they store? (Use k = 9.00 times 10^9 N middot m^2/C^2.) + 5.00 J +3240 J -5.00 J -3240 J How much work must have been done by the electric force to bring the +7.80 mu C point all the way from infinity to that location? (Use k = 9.00 x10^9 N middot m^2/C^2.) +5.00 J 3240 J -5.00 J -3240 J At the locations illustrated in the diagram: what is the voltage setup by the two charges at a point exactly between them? +4.85 times 10^6 V -3.46 times 10^6 V -2.33 times 10^6 V +2.29 times 10^6 V With the two charges held at their positions, as illustrated: If now a third point charge +6.40 mu C were to be brought from infinitely far away to the point exactly between the charges, how much work would the electric force do?

Explanation / Answer

13)

here,

charge on proton , q = 1.6 * 10^-19 C

sepration , d = 2 * 10^-10 m

the electric potential of the charges , E = k * q^2 /d

E = 9 * 10^9 * (1.6 * 10^-19)^2 /(2 * 10^-10) J

E = 1.15 * 10^-18 J

the electric potential energy of the charges is b) 1.15 * 10^-18 J

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