Show that the electric potential of carbon nucleus of equivalent charge +6e at a distance of 1x10^-10 m from a carbon nucleus with constant k of 9 x 10^9 is 86.4V
How do I answer this question?
V = k Q/r where Q = 6* 1.6*10^-19 Coulombs
= 9*10^9 (6)(1.6*10^-19)/10^-10
= 86.4 * 10^(9+10-19)
= 86.4
remember E = k Q /r^2
V = -integral of E from infinity to r= work done to bring 1 C charge in from infinity , force opposite to motion
V = -k Q (-1/r) ......at r - at infinity
= -k Q [ -1/r +1/infinity]
= k Q / r
To find the electric potential of the carbon nucleus at a given distance, you need to use the formula for electric potential due to a point charge. Here's how you can solve this problem:
1. Determine the value of the electric potential due to a point charge using the formula: V = k * (q / r), where V is the electric potential, k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge.
2. Calculate the electric potential due to the carbon nucleus of equivalent charge +6e using the given charge value. Since e represents the elementary charge, +6e is equivalent to 6 times the elementary charge (+6e = +6 * 1.6 x 10^-19 C).
3. Plug in the known values into the formula: V = (9 x 10^9 Nm^2/C^2) * ((6 * 1.6 x 10^-19 C) / (1 x 10^-10 m)). Simplify the equation.
4. Calculate the final answer to get the electric potential.
By performing these calculations, you will find that the electric potential of the carbon nucleus of equivalent charge +6e at a distance of 1x10^-10 m from a carbon nucleus is 86.4V.