Three point charges are placed on the x-y plane as follows: Q1=-1.2nC at ( 3,4)cm, Q2=2.3nC at (0,0)cm and Q3=-3.7nC at (5,0)cm. What is the net force exerted by the two point charges on Q2?
F = k Q1 Q2/d^2
I am ignoring powers of ten and k
distance of Q1 from Q2 = 5 (hypotenuse)
cos angle = 3/5
sin angle = 4/5
in x direction:
2.3 * 2 * (3/5)/25 + 2.3 * 3.7 /25
note that 25 is cm^2, multiply by 10^-4
for m^2
in y direction:
2.3 * 2 * (4/5)/25
magnitude = (Fx^2+Fy^2)^.5
tan angle = Fy/Fx
8.75x10^-5,65.3 degrees above +x axis
To find the net force exerted by the two point charges on Q2, we need to calculate the force exerted by each individual charge and then sum them up.
The formula to calculate the force between two charges is given by Coulomb's Law:
F = (k * |Q1 * Q2|) / r^2
Where:
- F is the force between the charges
- k is the Coulomb's constant (k = 9 * 10^9 N m^2/C^2)
- Q1 and Q2 are the magnitudes of the two charges
- r is the distance between the charges
Let's calculate the force exerted by Q1 on Q2:
Q1 = -1.2nC = -1.2 * 10^-9 C
Q2 = 2.3nC = 2.3 * 10^-9 C
r = √((3-0)^2 + (4-0)^2) = √(9 + 16) = √25 = 5 cm = 5 * 10^-2 m
Plugging these values into the formula, we get:
F1 = (9 * 10^9 N m^2/C^2) * |-1.2 * 10^-9 C * 2.3 * 10^-9 C| / (5 * 10^-2 m)^2
Calculating this expression, we find:
F1 ≈ 2.068 * 10^-5 N
Now, let's calculate the force exerted by Q3 on Q2:
Q3 = -3.7nC = -3.7 * 10^-9 C
r = √((5-0)^2 + (0-0)^2) = √25 = 5 cm = 5 * 10^-2 m
Plugging these values into the formula, we get:
F3 = (9 * 10^9 N m^2/C^2) * |-3.7 * 10^-9 C * 2.3 * 10^-9 C| / (5 * 10^-2 m)^2
Calculating this expression, we find:
F3 ≈ 2.266 * 10^-5 N
Finally, we can calculate the net force by summing up the forces:
Net Force = F1 + F3
Net Force ≈ 2.068 * 10^-5 N + 2.266 * 10^-5 N
Adding these values, we find:
Net Force ≈ 4.334 * 10^-5 N
Therefore, the net force exerted by the two point charges on Q2 is approximately 4.334 * 10^-5 N.
To find the net force exerted on Q2 by the other two point charges, we can use Coulomb's Law, which states that the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:
F = k * |Q1 * Q2| / r^2
Where:
F is the force between two charges
k is Coulomb's constant (k ≈ 8.99 x 10^9 Nm^2/C^2)
|Q1 * Q2| is the product of the magnitudes of the charges
r is the distance between the charges.
In this case, we want to find the net force on Q2, so we need to calculate the force exerted by Q1 and Q3 separately, and then find the vector sum of the two forces.
Let's calculate the force exerted by Q1 on Q2 first.
Q1 = -1.2nC (negative sign indicates that it has an opposite charge compared to Q2)
Q2 = 2.3nC
r = distance between Q1 and Q2
r = √((x2 - x1)^2 + (y2 - y1)^2)
Using the coordinates provided:
r = √((0 - 3)^2 + (0 - 4)^2)
r ≈ √((-3)^2 + (-4)^2)
r ≈ √(9 + 16)
r ≈ √25
r = 5 cm
Now, let's calculate the force exerted by Q1 on Q2:
F1 = k * |Q1 * Q2| / r^2
F1 = (8.99 x 10^9 Nm^2/C^2) * |-1.2nC * 2.3nC| / (5 cm)^2
|-1.2nC * 2.3nC| ≈ 2.76 x 10^-9 C^2 (product of the magnitudes of the charges)
F1 ≈ (8.99 x 10^9 Nm^2/C^2) * (2.76 x 10^-9 C^2) / (5 x 10^-2 m)^2
F1 ≈ (8.99 x 2.76) * 10^(-9 -2 -2) N
F1 ≈ 24.8024 x 10^-9 N
F1 ≈ 2.48024 x 10^-8 N
Next, let's calculate the force exerted by Q3 on Q2.
Q3 = -3.7nC (negative sign indicates that it has an opposite charge compared to Q2)
Q2 = 2.3nC
r = distance between Q3 and Q2
r = √((x2 - x3)^2 + (y2 - y3)^2)
Using the coordinates provided:
r = √((0 - 5)^2 + (0 - 0)^2)
r ≈ √((-5)^2 + 0^2)
r ≈ √(25)
r = 5 cm
Now, let's calculate the force exerted by Q3 on Q2:
F3 = k * |Q3 * Q2| / r^2
F3 = (8.99 x 10^9 Nm^2/C^2) * |-3.7nC * 2.3nC| / (5 cm)^2
|-3.7nC * 2.3nC| ≈ 8.51 x 10^-9 C^2 (product of the magnitudes of the charges)
F3 ≈ (8.99 x 10^9 Nm^2/C^2) * (8.51 x 10^-9 C^2) / (5 x 10^-2 m)^2
F3 ≈ (8.99 x 8.51) * 10^(-9 -2 -2) N
F3 ≈ 76.5449 x 10^-9 N
F3 ≈ 7.65449 x 10^-8 N
Now, to find the net force on Q2, we need to add the forces F1 and F3 as vectors.
Net force on Q2 = F1 + F3
Net force ≈ (2.48024 x 10^-8 N) + (7.65449 x 10^-8 N)
Net force ≈ 10.13473 x 10^-8 N
Therefore, the net force exerted by the two point charges on Q2 is approximately 10.13473 x 10^-8 N.