At what distance from a carbon nucleus is the electric potential 2.1 V? Carbon atoms have 6 electrons in the atomic shell.
use voltage potential for point charge
v= K Q / r^2
V given, K known, Q ( 6* electron charge) , solve for r
V=kQ/r
not V=KQ/r^2
To determine the distance from a carbon nucleus where the electric potential is 2.1 V, we need to consider the concept of electric potential due to point charges.
The electric potential due to a point charge can be calculated using the equation:
V = k * (Q / r)
Where:
- V is the electric potential,
- k is the electrostatic constant (approximately equal to 9 × 10^9 Nm^2/C^2),
- Q is the charge of the point source, and
- r is the distance from the point charge.
In the case of a carbon nucleus, it is important to note that the electric potential is created by the 6 electrons located in the atomic shell. Each electron has a charge of -1.6 × 10^-19 C.
Since we are given the value of the electric potential (2.1 V), we can rearrange the equation to solve for the distance (r):
r = k * (Q / V)
Substituting the values:
Q = (-1.6 × 10^-19 C) * 6, as there are 6 electrons in a carbon atom.
V = 2.1 V
k = 9 × 10^9 Nm^2/C^2
Plugging in these values:
r = (9 × 10^9 Nm^2/C^2) * ((-1.6 × 10^-19 C) * 6) / (2.1 V)
Calculating the above expression will give us the required distance from the carbon nucleus where the electric potential is 2.1 V.