The distance between two charges q1 = -2.5 μC and q2 = 6.0 μC is equal to 1.0 m. Determine the point (other than infinity) at which the electric field is zero.
If we place the -2.5 charge at zero and 6.0 charge at 1 the E field will be zero somewhere (distance x) on the negative x axis.
so k q1/x(sqr) = k q2/(x+1)sqr or
q2/q1 = [(x+1)sqr]/x(sqr)= (x+1/x)sqr
so 2.4 = (x+1/x)sqr
1.55 = x+1/x = 1 + 1/x
.55 = 1/x
x = 1.82m
To find the point at which the electric field is zero, we can use the principle of superposition. The electric field at a point due to multiple charges is the vector sum of the electric fields produced by each individual charge.
Let's assume the position of charge q1 is at the origin (0,0) and the position of charge q2 is at (1.0 m, 0).
The formula to calculate the electric field at a given point due to a point charge is:
E = k * (q / r^2)
Where:
E is the electric field,
k is the electrostatic constant (9.0 x 10^9 N·m^2/C^2),
q is the charge,
r is the distance between the point and the charge.
For charge q1:
E1 = k * (q1 / r1^2) = -2.5 * 10^-6 C / r1^2
For charge q2:
E2 = k * (q2 / r2^2) = 6.0 * 10^-6 C / r2^2
To find the point where the electric field is zero, we need to set E1 + E2 = 0.
Therefore, -2.5 * 10^-6 C / r1^2 + 6.0 * 10^-6 C / r2^2 = 0
Simplifying this equation, we get:
2.5 / r1^2 = 6.0 / r2^2
r2^2 / r1^2 = 2.4
Taking the square root of both sides, we get:
r2 / r1 = √2.4
r2 = √2.4 * r1
Now, we substitute the coordinates for r1 and r2:
r1 = 0 (since we assumed q1 is at the origin)
r2 = √2.4 * 1.0 m
r2 = √2.4 m
Therefore, the point at which the electric field is zero (other than infinity) is (√2.4 m, 0).
To determine the point at which the electric field is zero, we can calculate the distance x from charge q1 where the electric field from q1 is equal in magnitude but opposite in direction to the electric field from q2.
The formula for the electric field from a point charge is given by:
E = k * q / r^2
Where E is the electric field, 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.
For charge q1, the electric field at distance x is:
E1 = k * q1 / (x^2)
For charge q2, the electric field at distance (1 - x) is:
E2 = k * q2 / ((1 - x)^2)
Since the electric field should be zero, we can set these two equations equal to each other:
E1 = E2
k * q1 / (x^2) = k * q2 / ((1 - x)^2)
Now, substitute the given values:
(-2.5 μC) / (x^2) = (6.0 μC) / ((1 - x)^2)
To simplify the calculation, let's convert the charges to coulombs:
(-2.5 μC) = (-2.5 x 10^-6 C)
(6.0 μC) = (6.0 x 10^-6 C)
(-2.5 x 10^-6) / (x^2) = (6.0 x 10^-6) / ((1 - x)^2)
Cross-multiply:
(-2.5 x 10^-6) * ((1 - x)^2) = (6.0 x 10^-6) * (x^2)
Expand the brackets:
(-2.5 x 10^-6) * (1 - 2x + x^2) = (6.0 x 10^-6) * (x^2)
Distribute the terms:
-2.5 x 10^-6 + 5 x 10^-6 x - 2.5 x 10^-6 x^2 = 6.0 x 10^-6 x^2
Combine like terms:
2.5 x 10^-6 x^2 + 5 x 10^-6 x - 6.0 x 10^-6 x^2 = 0
4.5 x 10^-6 x^2 + 5 x 10^-6 x = 0
Factor out x:
x * (4.5 x 10^-6 x + 5 x 10^-6) = 0
Now we have two possible solutions:
1) x = 0
2) 4.5 x 10^-6 x + 5 x 10^-6 = 0
For x = 0, the electric field is zero at the origin.
For 4.5 x 10^-6 x + 5 x 10^-6 = 0, we solve for x:
4.5 x 10^-6 x + 5 x 10^-6 = 0
4.5 x 10^-6 x = -5 x 10^-6
x = -(5 x 10^-6) / (4.5 x 10^-6)
x = -5/4.5
x = -1.11 m
Therefore, there are two points (other than infinity) at which the electric field is zero: x = 0 (origin) and x = -1.11 m (negative direction from charge q1).