A proton has a positive electric charge of q = 1.6 × 10–19 coulombs. What is the electric potential at a point 5.3 × 10–11 m from the proton?


4.4 * 10 ^-18

2.7 * 10 ^1

3.0 * 10 ^-9

5.1 * 10 ^1

Respuesta :

2.7 × 101 volts is the correct awnser

Answer: 27.1 Volts.

The formula of potential difference is,

[tex] V = \frac{kq}{r} [/tex]

The charge on the proton is, [tex] q = 1.6 \times 10^{-19} C [/tex].

The distance of the point of observation from proton is, [tex] r = 5.3 \times 10^{-11} m [/tex]

The coulomb's constant, [tex] k = 9 \times 10^9 V/C.m [/tex]

The potential is,

[tex] V = \frac{9 \times 10^9 Vm/C \times 1.6 \times 10^{-19} C}{5.3 \times 10^{-11} m}
=27.1 V [/tex]

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