1. three point charges, q1, q2, q3, lie along the x-axis at x = 0,3,5, respectively. Calculate the magnitude and direction of the net electric force on q1 if q1 = +6 x 10^-6 C, q2 = +1.5 X 10^-6 C and q3 = -2.0 x 10^-6 C.
my answre is 46.8 N to the right. is that correct?
2. a charge of +2.0 x 10^-3 C is placed at the origin, and antoher charge of +4.0 x 10^-3 C is placed at x = 2m.
if a test charge, q, of +3 x 10^-9 C is placed at 1m, then determine the net electric field at that point.
my answer is 5.4 x 10^7 N/C
is that correct
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1. In problem #1, you do not say if the locations 0, 3 and 5 are in meters or some other units. That has to be specified for a numerical answer to be obtained.
At location x=0 there is a push to the left dye to the q2-q1 repulsion and a pull to the right due to the q1-q3 attraction.
I the R distances are in meters, t he two forces are k q1q2/3^2 and k q1 q3/5^2], where k = 8.99*10^9 N/m^2C^2
This leads to a much smaller answer than yours.
2. The field at the test charge position has nothing to do with the amount of the test charge.
The E-field is k(2*10^-3)/1^2 - k(4*10^-3)/1^2
The sign changes is because charges are pushing from opposite sides. It appears you added the fields from the two charges instead iof taking the difference.
Let's start with the first question:
To find the net electric force on q1, we need to calculate the force due to each individual charge and then add them up.
The electric force between two point charges is given by Coulomb's Law:
F = k * |q1 * q2| / r^2
Where F is the magnitude of the electric force, k is the electrostatic constant (k ≈ 9x10^9 N * m^2 / C^2), q1 and q2 are the charges, and r is the distance between the charges.
First, let's calculate the force between q1 and q2:
F12 = (9x10^9) * |(6x10^-6) * (1.5x10^-6)| / (3-0)^2
= (9x10^9) * (9x10^-12) / 9
= 9x10^-3 N to the right
Next, let's calculate the force between q1 and q3:
F13 = (9x10^9) * |(6x10^-6) * (-2x10^-6)| / (5-0)^2
= (9x10^9) * (12x10^-12) / 25
= 4.32x10^-3 N to the left
Finally, let's calculate the net force on q1 by subtracting F13 from F12:
Net force = F12 - F13
= 9x10^-3 N to the right - 4.32x10^-3 N to the left
= 4.68x10^-3 N to the right
So, the magnitude of the net electric force on q1 is 4.68x10^-3 N to the right.
Now let's move on to the second question:
To find the net electric field at a point, we need to calculate the electric field due to each individual charge and then add them up.
The electric field due to a point charge is given by the formula:
E = k * |q| / r^2
Where E is the magnitude of the electric field, k is the electrostatic constant (k ≈ 9x10^9 N * m^2 / C^2), q is the charge, and r is the distance from the charge.
First, let's calculate the electric field due to the charge at the origin:
E1 = (9x10^9) * |2x10^-3| / (1-0)^2
= (9x10^9) * (2x10^-3) / 1
= 18x10^6 N/C away from the origin
Next, let's calculate the electric field due to the charge at x = 2m:
E2 = (9x10^9) * |4x10^-3| / (1-2)^2
= (9x10^9) * (4x10^-3) / 4
= 9x10^6 N/C towards the origin
Now, let's calculate the net electric field by adding up the individual electric fields:
Net electric field = E1 + E2
= 18x10^6 N/C away from the origin + 9x10^6 N/C towards the origin
= 27x10^6 N/C away from the origin
So, the net electric field at the point is 27x10^6 N/C away from the origin.
Let's start by solving the first question. We have three point charges located along the x-axis with their respective positions and magnitudes given as follows:
Charge q1: +6 x 10^-6 C (located at x = 0)
Charge q2: +1.5 x 10^-6 C (located at x = 3)
Charge q3: -2.0 x 10^-6 C (located at x = 5)
To find the net electric force on q1, we can use the concept of Coulomb's Law which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as:
F = k * |q1*q2| / r^2
where,
F is the magnitude of the force,
k is the electrostatic constant (k = 9 x 10^9 Nm^2/C^2),
|q1*q2| is the product of the magnitudes of the two charges, and
r is the distance between them.
Now let's calculate the force between q1 and q2:
F1-2 = (9 x 10^9 Nm^2/C^2) * |(+6 x 10^-6 C) * (+1.5 x 10^-6 C)| / (3 - 0)^2
F1-2 = (9 x 10^9 Nm^2/C^2) * (9 x 10^-12 C^2) / 9
F1-2 = 9 N
Next, let's calculate the force between q1 and q3:
F1-3 = (9 x 10^9 Nm^2/C^2) * |(+6 x 10^-6 C) * (-2.0 x 10^-6 C)| / (5 - 0)^2
F1-3 = (9 x 10^9 Nm^2/C^2) * (12 x 10^-12 C^2) / 25
F1-3 = -4.32 N
Since q3 is negative, the force is attractive and its direction is towards q3.
Now, to find the net force on q1, we need to sum the forces F1-2 and F1-3:
Net force on q1 = F1-2 + F1-3
= 9 N + (-4.32 N)
= 4.68 N (rounded to two decimal places)
The direction of the net electric force on q1 is given by the vector sum of the individual forces. Since both F1-2 and F1-3 are acting in the positive x-direction, the net force will also be in the positive x-direction (to the right). Therefore, your answer of 46.8 N to the right seems incorrect. The correct answer is 4.68 N to the right.
Now, let's move on to the second question.
We have two point charges located on the x-axis as follows:
Charge Q1: +2.0 x 10^-3 C (located at the origin, x = 0)
Charge Q2: +4.0 x 10^-3 C (located at x = 2 m)
To determine the net electric field at a point located at x = 1 m, we can again use Coulomb's Law and equation for electric field intensity. For a charge q at a point P, the electric field intensity at that point is given by:
E = k * |Q| / r^2
where,
E is the magnitude of the electric field intensity,
k is the electrostatic constant (k = 9 x 10^9 Nm^2/C^2),
|Q| is the magnitude of the charge, and
r is the distance between the charge and the point where electric field is to be calculated.
First, let's calculate the electric field at point P (x = 1m) due to Q1:
E1 = (9 x 10^9 Nm^2/C^2) * (|+2.0 x 10^-3 C|) / (1)^2
E1 = (9 x 10^9 Nm^2/C^2) * (2.0 x 10^-3 C) / 1
E1 = 18 x 10^6 N/C
Next, let's calculate the electric field at point P (x = 1m) due to Q2:
E2 = (9 x 10^9 Nm^2/C^2) * (|+4.0 x 10^-3 C|) / (2 - 1)^2
E2 = (9 x 10^9 Nm^2/C^2) * (4.0 x 10^-3 C) / 1
E2 = 36 x 10^6 N/C
Now, to find the net electric field at point P, we need to sum the electric fields E1 and E2:
Net electric field at point P = E1 + E2
= 18 x 10^6 N/C + 36 x 10^6 N/C
= 54 x 10^6 N/C
= 5.4 x 10^7 N/C (in scientific notation)
Hence, your answer of 5.4 x 10^7 N/C is correct.