Two point charges Q1 = 10µc and Q2 = 2µc are arranged along the x-axis at
x = 0 and x=4m respectively. Find the position along the x-axis where v = 0.
To find the position along the x-axis where the electric potential (v) is equal to zero, we need to consider the electric potential due to both charges separately and then find the location where their potentials cancel each other out.
The electric potential (V) due to a point charge Q at a distance r from it is given by the equation:
V = k(Q/r)
where k is the electrostatic constant (k = 9 x 10^9 Nm^2/C^2).
Let's calculate the electric potentials due to each charge separately:
For Q1 at x = 0:
V1 = k(Q1/r1)
For Q2 at x = 4m:
V2 = k(Q2/r2)
Since we want to find the position where V = 0, we can write:
V1 - V2 = 0
k(Q1/r1) - k(Q2/r2) = 0
Dividing both sides by k:
(Q1/r1) - (Q2/r2) = 0
Simplifying further:
(Q1/r1) = (Q2/r2)
Substituting the values of the charges:
(10µC/r1) = (2µC/r2)
Now, we know that r1 + r2 = 4m (total distance along the x-axis).
We can solve these two equations to find the values of r1 and r2.
Let's substitute r2 = (4m - r1) in the equation:
(10µC/r1) = (2µC/(4m - r1))
Now cross multiply:
10µC * (4m - r1) = 2µC * r1
(40µC - 10µCr1) = 2µCr1
40µC = 2µCr1 + 10µCr1
40µC = 12µCr1
Dividing both sides by 12µC:
r1 = 40µC / 12µC
r1 = 3.33 m (rounded to two decimal places)
Therefore, the position along the x-axis where V = 0 is approximately 3.33 meters from Q1 (at x = 0) and 0.67 meters from Q2 (at x = 4m).
To find the position along the x-axis where the electric potential (V) is equal to zero (v = 0) due to two point charges Q1 and Q2, we need to calculate the electric potential at different points along the x-axis and find the point where it equals zero.
The electric potential (V) at any point due to a point charge Q can be calculated using the formula:
V = k * Q / r
where V is the electric potential, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the point charge, and r is the distance between the point charge and the point where we want to calculate the electric potential.
In this case, we have two point charges, Q1 and Q2, arranged along the x-axis at x = 0 and x = 4m, respectively.
Let's calculate the electric potential due to each charge at different points along the x-axis:
1. Electric potential due to Q1 at any point x on the x-axis:
V1 = k * Q1 / (x - 0)
2. Electric potential due to Q2 at any point x on the x-axis:
V2 = k * Q2 / (x - 4)
To find the position where v = 0, we need to solve the equation:
V1 + V2 = 0
Substituting the values of V1 and V2 into the equation, we get:
(k * Q1 / x) + (k * Q2 / (x - 4)) = 0
Now, let's substitute the given values into the equation:
Q1 = 10µC (10 x 10^-6 C)
Q2 = 2µC (2 x 10^-6 C)
k = 9 x 10^9 Nm^2/C^2
(k * (10 x 10^-6) / x) + (k * (2 x 10^-6) / (x - 4)) = 0
Simplifying the equation, we get:
(10 * k) / x + (2 * k) / (x - 4) = 0
Now, we can solve this equation to find the position (x) where v = 0.
field strength is proportional to the charge
... and inversely proportional to the square of the distance
d is the distance from the origin along the x-axis
10 / d^2 = 2 / (4 - d)^2
10 (4 - d^2) = 2 d^2
40 = 12 d^2