In (Figure 1), the charge q=1.28×10^−9 C.
figure contain ( -2q at 0 and +5q at 1m )
a) Find the electric potential at x=0.750 m.(include appropriate units)
b) Find the value of x between 0 and 1.00 m where the electric potential is zero. (include appropriate units)
To find the electric potential at a specific point, we can use the formula:
V = kq / r
where V is the electric potential, k is the Coulomb's constant (9 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 = 0.750 m, we need to consider the contributions from both charges -2q and +5q.
For -2q at x = 0, the distance from the charge is:
r1 = 0.750 m
Plugging in the values, the electric potential due to -2q at x = 0 is:
V1 = (9 x 10^9 N m^2/C^2) * (-2q) / (r1)
For +5q at x = 1 m, the distance from the charge is:
r2 = 1 m - 0.750 m = 0.250 m
Plugging in the values, the electric potential due to +5q at x = 1 m is:
V2 = (9 x 10^9 N m^2/C^2) * (+5q) / (r2)
The total electric potential at x = 0.750 m is obtained by adding the contributions from both charges:
V_total = V1 + V2
Remember to include the appropriate units in the final answer.
b) To find the value of x between 0 and 1.00 m where the electric potential is zero, we need to set up the equation:
V_total = 0
Using the same approach as before, we can calculate the electric potentials due to each charge at an arbitrary distance x:
V1 = (9 x 10^9 N m^2/C^2) * (-2q) / x
V2 = (9 x 10^9 N m^2/C^2) * (+5q) / (1 - x)
Solving the equation V_total = V1 + V2 = 0 for x will give us the value where the electric potential is zero.
To find the electric potential at a certain point due to the given charges, we can use the formula:
V = k * (q1 / r1 + q2 / r2 + q3 / r3 + ...)
where V is the electric potential, k is the electrostatic constant (9 x 10^9 N m^2 / C^2), q1, q2, etc. are the charges, and r1, r2, etc. are the distances from the charges to the point.
a) To find the electric potential at x = 0.750 m, we need to calculate the contribution from each charge to the total electric potential.
Given that q = 1.28 × 10^−9 C, the charges are -2q (located at 0 m) and +5q (located at 1 m).
1. Contribution from -2q (located at 0 m):
distance (r1) = |0.750 - 0| = 0.750 m
potential (V1) = k * (-2q / r1)
2. Contribution from +5q (located at 1 m):
distance (r2) = |0.750 - 1| = 0.250 m
potential (V2) = k * (5q / r2)
Now, let's substitute the given values into the formula and calculate the electric potential at x = 0.750 m:
V = k * (-2q / r1) + k * (5q / r2)
Remember to use the proper values for k and q.
b) To find the value of x between 0 and 1.00 m where the electric potential is zero, we need to set the total electric potential formula to zero and solve for x.
Using the same formula as above:
V = k * (-2q / r1) + k * (5q / r2) = 0
Now, you can solve this equation for x. Keep in mind that x should lie between 0 and 1.00 m.
I have no idea what the figure shows, my crystal ball is broken. On the line "figure contain", is that on the x axis, or y axis?
If it is on the x axis, notice that in between x=0 and x=1m, the two forces oppose, so in your calcualtions, watch sign. So the potentials due to those charges are opposite signs.
Elctric potential at x = kq(-2q/x^2 + 5q/(1-x)^2)