A point charge has an excess of 5.0 × 1012 electrons. What would be the electric potential at a distance of 0.50 m from the charge? (e = 1.6 × 10-19 C)
Givens: N = 5.0 x 10^12, r = 0.50 m, e = 1.6 x 10^-19
First use the equation q = Ne to find the charge.
q = (5.0 x 10^12) x (-1.6 x 10^-19)
q = -8.0 x 10^-7
The elementary charge here is “negative” because there is an excess of electrons. If the object had lost electrons, the balloon would have a deficit of electrons, e.g., more protons than electrons. The elementary charge would then be “positive”.
Then use the equation V = kq/r to find electric potential
V = (9 x 10^9) x (-8.0 x 10^-7) / 0.50 m
V = -1.44 x 10^4 V
wdym
nvm thanks
can you help me out I don't know what you mean in the text
To find the electric potential at a distance from a point charge, you can use the formula:
V = ke * q / r
Where:
V is the electric potential,
ke is the electrostatic constant (equal to 9 × 10^9 N m^2/C^2),
q is the charge, and
r is the distance from the charge.
First, let's find the charge of the point charge. We're told that it has an excess of 5.0 × 10^12 electrons. The charge of a single electron is -1.6 × 10^-19 C. So, the total charge of the point charge would be:
q = (5.0 × 10^12) * (-1.6 × 10^-19)
= -8.0 × 10^-7 C
Now we can calculate the electric potential. Given that the distance from the charge is 0.50 m:
V = (9 × 10^9) * (-8.0 × 10^-7) / (0.50)
= -1.44 × 10^5 V
Therefore, the electric potential at a distance of 0.50 m from the point charge would be -1.44 × 10^5 V.