4.
Three point charges, +6.3 µC, +2.3 µC, and
−3.4 µC, lie along the x-axis at 0 cm, 2.6 cm,
and 5.2 cm, respectively.
What is the force exerted on q1 by the other
two charges? (To the right is positive.) The
Coulomb constant is 8.99 × 109 N · m
2
/C
2
.
Answer in units of N.
5.
What is the force exerted on q2 by the other
two charges? (To the right is positive.)
Answer in units of N
My work for 4.
Fa=8.99E9 x 6.6E-6 x 2.3E-6 / 0.026^2
Fa=192.68998521
Fb=8.99E9 x 6.3e-6 x 3.4E-6/0.052^2
Fb=71.21516272
Fa+Fb= -121
My work for 5.
Fa=8.99E9 x 2.3E-6 x 6.3E-6/0.026^2
Fa=192.9142012
Fb=8.99e9 x 2.3E-6 x 3.4E-6/0.052^2
Fb=-26
192-26 = 166.8860947?????
Number 4. is correct but I know number 5. is not!
Answer:
4. -121 N
5. -26 N
Ah, I see what you did there. You tried to calculate the forces using the formula, but it seems like there may have been a little mix-up in the signs. Let me help you straighten things out with a little humor!
For question 4, to find the force on q1, you correctly used the formula:
Fa = k(q1q2/r^2)
But the result you got, -121 N, seems a bit negative. Maybe it's feeling a bit down today? Well, don't worry! Let's flip its mood by taking the absolute value. So the force on q1 is actually 121 N. Don't be so negative, q1!
Now, let's move on to the tricky one, question 5. You calculated the forces correctly but seemed to have a sign mix-up. The force on q2 is actually:
Fa = 8.99E9 * 2.3E-6 * (-6.3E-6) / (0.026)^2
Fa ≈ -174.89 N
Fb = 8.99E9 * 2.3E-6 * (-3.4E-6) / (0.052)^2
Fb ≈ 31.16 N
Now, when we add these forces together, we get:
(-174.89 N) + (31.16 N) ≈ -143.73 N
So, the force on q2 by the other two charges is approximately -143.73 N. Looks like q2 might need a little cheering up too!
Remember, when dealing with forces, positive and negative signs can really change the mood. Keep that in mind and keep up the good work!
To calculate the force exerted on q1 by the other two charges, we can use the formula for the electric force between two point charges:
F = k * |q1 * q2| / r^2
where F is the force, k is the Coulomb constant (8.99 × 10^9 N · m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
For q1 being influenced by the other two charges (+2.3 µC and -3.4 µC), we can calculate the force as follows:
F1 = k * |q1 * q2| / r1^2
= (8.99 × 10^9 N · m^2/C^2) * |(6.3 µC) * (-3.4 µC)| / (0.026 m)^2
= (8.99 × 10^9) * (6.3 * 10^(-6)) * (3.4 * 10^(-6)) / (0.026)^2
= 193.848 N
Therefore, the force exerted on q1 by the other two charges is 193.848 N.
For calculating the force exerted on q2 by the other two charges, we can follow the same approach:
F2 = k * |q2 * q3| / r2^2
= (8.99 × 10^9 N · m^2/C^2) * |(2.3 µC) * (-3.4 µC)| / (0.052 m)^2
= (8.99 × 10^9) * (2.3 * 10^(-6)) * (3.4 * 10^(-6)) / (0.052)^2
= 45.11 N
Therefore, the force exerted on q2 by the other two charges is 45.11 N.
To calculate the force exerted on q1 by the other two charges, we need to use Coulomb's law. Coulomb's law states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for calculating the force is:
F = k * (q1 * q2) / r^2
Where:
F is the force between the charges,
k is the Coulomb constant (8.99 × 10^9 N·m^2/C^2),
q1 and q2 are the charges of the two point charges, and
r is the distance between the charges.
For number 4:
- Charge q1 is +6.3 µC, charge q2 is +2.3 µC, and charge q3 is -3.4 µC.
- The distance between q1 and q2 is 2.6 cm (0.026 m), and between q1 and q3 is 5.2 cm (0.052 m).
To calculate the force on q1 due to q2 (F_q1q2), we use the formula:
F_q1q2 = k * (q1 * q2) / r^2
F_q1q2 = 8.99 × 10^9 N·m^2/C^2 * (+6.3 µC * +2.3 µC) / (0.026 m)^2
Simplifying the calculation gives:
F_q1q2 = 192.69 N
To calculate the force on q1 due to q3 (F_q1q3), we use the same formula:
F_q1q3 = k * (q1 * q3) / r^2
F_q1q3 = 8.99 × 10^9 N·m^2/C^2 * (+6.3 µC * -3.4 µC) / (0.052 m)^2
Simplifying the calculation gives:
F_q1q3 = -121 N
Therefore, the total force exerted on q1 by the other two charges is:
F_total = F_q1q2 + F_q1q3
F_total = 192.69 N + (-121 N)
F_total = 71.69 N (rounded to two decimal places)
For number 5:
To calculate the force exerted on q2 by the other two charges, we follow the same steps as before and use the formula:
F_q2q1 = k * (q2 * q1) / r^2
F_q2q1 = 8.99 × 10^9 N·m^2/C^2 * (+2.3 µC * +6.3 µC) / (0.026 m)^2
Simplifying the calculation gives:
F_q2q1 = 192.69 N
To calculate the force on q2 due to q3 (F_q2q3), we use the formula:
F_q2q3 = k * (q2 * q3) / r^2
F_q2q3 = 8.99 × 10^9 N·m^2/C^2 * (+2.3 µC * -3.4 µC) / (0.052 m)^2
Simplifying the calculation gives:
F_q2q3 = -28 N (rounded to two decimal places)
Therefore, the total force exerted on q2 by the other two charges is:
F_total = F_q2q1 + F_q2q3
F_total = 192.69 N + (-28 N)
F_total = 164.69 N (rounded to two decimal places)
So, the correct answer for number 5 is 164.69 N, not 166.89 N as previously mentioned.