4. A positive point charge q=+2.5-nC is located at x=1.20 m and a negative charge of -2q=- 5.00-nC is located at the origin as shown in Figure below
a. Find the electric potential at x=0.600 m
b. Find the point along the x-axis between the two charges where the electric potential is zero.
To find the electric potential at a specific point, we need to use the formula for the electric potential:
V = k * Q / r
where V is the electric potential, k is the electrostatic constant (8.99 x 10^9 N m^2/C^2), Q is the charge, and r is the distance from the charge.
a. To find the electric potential at x = 0.600 m, we need to calculate the electric potential due to both charges at this point.
For the positive charge at x = 1.20 m and Q = 2.5 nC:
V_pos = k * Q_pos / r_pos
where r_pos is the distance from the positive charge to the point of interest.
Given that r_pos = 0.600 m - 1.20 m = -0.600 m (negative because the positive charge is to the left of the point), plug in the values:
V_pos = (8.99 x 10^9 N m^2/C^2) * (2.5 x 10^-9 C) / (-0.600 m)
Similarly, for the negative charge at the origin and Q = -5.00 nC:
V_neg = k * Q_neg / r_neg
where r_neg is the distance from the negative charge to the point of interest.
Given that r_neg = 0.600 m (since the negative charge is at the origin and the point of interest is to the right), plug in the values:
V_neg = (8.99 x 10^9 N m^2/C^2) * (-5.00 x 10^-9 C) / (0.600 m)
To find the total electric potential at x = 0.600 m, subtract V_neg from V_pos:
V_total = V_pos + V_neg
b. To find the point along the x-axis between the two charges where the electric potential is zero, we can set the total electric potential to zero:
V_total = 0
Using the same formula for V_total as before, set it to zero and rearrange the equation to solve for x:
0 = (8.99 x 10^9 N m^2/C^2) * (2.5 x 10^-9 C) / (x - 1.20 m) + (8.99 x 10^9 N m^2/C^2) * (-5.00 x 10^-9 C) / x
Solve this equation for x using algebraic methods to find the point along the x-axis where the electric potential is zero.