There are 3 charges situated on the x-axis: a +3.0 nC charge at (-3 cm, 0), a +4.0 nC charge at the origin (0, 0) and a -6.0 nC charge at (3 cm, 0). What is the electric potential at the point (0, 3 cm) on the y-axis?
To find the electric potential at the point (0, 3 cm) on the y-axis, you can use the principle of superposition. The electric potential due to each individual charge is given by the equation:
V = k * q / r
Where:
V is the electric potential
k is the electrostatic constant (8.99 x 10^9 Nm^2/C^2)
q is the charge
r is the distance from the charge to the point of interest
In this case, we have three charges, so we need to calculate the electric potential due to each charge and sum them up.
Let's calculate the electric potential due to each charge:
1. +3.0 nC charge at (-3 cm, 0):
r1 = distance from point (0, 3 cm) to (-3 cm, 0) = √((0 - (-0.03))^2 + (0 - 0.03)^2) = 0.0424 m
V1 = (k * q1) / r1
2. +4.0 nC charge at the origin (0, 0):
r2 = distance from point (0, 3 cm) to (0, 0) = 0.03 m
V2 = (k * q2) / r2
3. -6.0 nC charge at (3 cm, 0):
r3 = distance from point (0, 3 cm) to (0.03 m, 0) = √((0.03 - 0)^2 + (0.03 - 0)^2) = 0.0424 m
V3 = (k * q3) / r3
Now, we can sum up the electric potentials due to each charge:
V_total = V1 + V2 + V3
Finally, substitute the values into the equation and calculate the electric potential at the point (0, 3 cm) on the y-axis.