Two charges of magnitudes Q1 = -2μc and Q2 = 6μc lie along the x-axis at a distance 4m and 5m from the origin respectively. At what point from the origin does the net electrostatic field be zero.
lets put the first one at 0 and the second one at 1
the answer HAS TO BE left of zero ! (why ? :)
( right one pushes left, left one pulls right :)
only ratio of charges matter, call it one and three coulombs
right pull by left one = left push by right one
magnitude of right pull = 2/(0-x)^2
magnitude of left push = 6/(1+x)^2
so
2/x^2 = 6/(x^2 + 2 x + 1)
2 x^2 + 4 x + 2 = 6 x^2
4 x^2 - 4 x - 2 = 0
2 x^2 - 2 x - 1 = 0
x = -.366 or+ 1.366
I think it is on the left of zero so -.366
NOW move it all 4 units right because we did it for x = 0 and one instead of 4 and 5
4 - .366 = 3.634
of course they could have intended x = -4 and +5, it is hard to say
Ah, an electrifying question! Let's see if I can shed some light on it, shall we?
To find the point where the net electrostatic field is zero, we need to consider the electric field vectors created by each charge and their relative magnitudes and directions.
The electric field due to a point charge is given by the equation:
E = (k * Q) / r²
Where E is the electric field, k is Coulomb's constant (9 x 10^9 Nm²/C²), Q is the charge, and r is the distance from the charge.
Now, let's calculate the electric field due to each charge and sum them up:
For Q1 (-2μc at 4m from the origin):
E1 = (9 x 10^9 Nm²/C² * -2μc) / 4²
For Q2 (6μc at 5m from the origin):
E2 = (9 x 10^9 Nm²/C² * 6μc) / 5²
The sum of these electric fields should be zero:
E1 + E2 = 0
Now let's solve for the point where the net electric field is zero. I'm not really good at math, so hold onto your electric charge and let's trust the numbers.
To find the point where the net electrostatic field is zero, we need to find the position where the individual electric fields due to the charges Q1 and Q2 cancel each other out.
Let's consider the point P(x, 0) on the x-axis where we want to determine the net electric field.
The electric field due to Q1 at point P can be calculated using Coulomb's law:
E1 = (k * |Q1|) / r1²
Where k is the electrostatic constant (9 x 10^9 Nm^2/C^2), |Q1| is the magnitude of Q1, and r1 is the distance between Q1 and point P.
Similarly, the electric field due to Q2 at point P can be calculated using:
E2 = (k * |Q2|) / r2²
Where |Q2| is the magnitude of Q2, and r2 is the distance between Q2 and point P.
Since the net electric field is zero, the magnitudes of the electric fields due to Q1 and Q2 at point P must be equal and opposite:
E1 = -E2
Now, we can substitute the formulas for E1 and E2 and solve for x.
(k * |Q1|) / r1² = -(k * |Q2|) / r2²
Simplifying the equation:
(|Q1| * r2²) = -(|Q2| * r1²)
Substituting the given values:
(-2μC * (5m)²) = -(6μC * (4m)²)
(50μC * m²) = (96μC * m²)
Now we solve for x by rearranging the equation:
x = √[(50 * 5²) / 96]
x = √[(1250 / 96)]
x ≈ 3.18 meters (rounded to two decimal places)
So, the point where the net electrostatic field is zero is approximately 3.18 meters from the origin along the x-axis.
To find the point where the net electrostatic field is zero, we need to find the position where the electric fields due to the two charges cancel each other out.
The electric field due to a point charge is given by Coulomb's Law:
E = k * (Q / r^2)
where E is the electric field, k is the Coulomb constant (9 x 10^9 N m^2/C^2), Q is the charge, and r is the distance from the charge.
Let's consider the electric field at a general point P(x, 0) on the x-axis. The electric field due to Q1 at point P is:
E1 = k * (Q1 / (x - 4)^2)
The electric field due to Q2 at point P is:
E2 = k * (Q2 / (x - 5)^2)
To cancel each other out, the magnitudes of E1 and E2 must be equal. So, we can set up the following equation:
|E1| = |E2|
Since we are looking for the position where the net electric field is zero, both E1 and E2 must have opposite directions. Hence, we need to consider the signs of the charges.
Given that Q1 = -2μC and Q2 = 6μC, we can write:
E1 = k * (|-2| / (x - 4)^2) = k * (2 / (x - 4)^2)
E2 = k * (|6| / (x - 5)^2) = k * (6 / (x - 5)^2)
Setting E1 equal to -E2, we have:
2 / (x - 4)^2 = -6 / (x - 5)^2
Cross-multiplying, we get:
2 * (x - 5)^2 = -6 * (x - 4)^2
Expanding, we have:
2 * (x^2 - 10x + 25) = -6 * (x^2 - 8x + 16)
Simplifying the equation further, we get:
2x^2 - 20x + 50 = -6x^2 + 48x - 96
Combining like terms, we have:
8x^2 - 68x + 146 = 0
Now, we can solve this quadratic equation to find the value(s) of x that satisfy the equation. We can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 8, b = -68, and c = 146. Substituting these values into the formula:
x = (-(-68) ± √((-68)^2 - 4 * 8 * 146)) / (2 * 8)
Simplifying this expression, we get:
x = (68 ± √(4624 - 4672)) / 16
x = (68 ± √(-48)) / 16
Since the expression under the square root is negative, it means there are no real solutions for x. This implies that there is no point on the x-axis where the net electric field is zero.