I need help on the following questions, I am stuck and not sure on what to do, can someone work it out and or tell me steps to obtain the answer. Thank you
1) A point charge of -0.90 microC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A point charge q is fixed to each of the remaining corners. The net force acting on either of the charges q is zero. Find the magnitude of q.
- i keep getting 3.6 micro C
2) A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and increases its linear momentum to 8.6 x 10^-23 kg·m/s from 2.4 x 10^-23 kg·m/s in a time of 5.7 x 10^-6 s. What is the magnitude of the electric field?
3) A uniform electric field has a magnitude of 3.1 x 10^3 N/C. In a vacuum, a proton begins with a speed of 2.5 x 10^4 m/s and moves in the direction of this field. Find the speed of the proton after it has moved a distance of 1.5 mm.
Edit - Question One answer was found, only need help on question 2 and 3 now
1) A point charge of -0.90 microC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A point charge q is fixed to each of the remaining corners. The net force acting on either of the charges q is zero. Find the magnitude of q.
- i keep getting 3.6 micro C
Find the repulsive force from the opposite corner. Then find the repulsive attractive force from one q IN THE DIRECTION of the diagonal kq(-.90)/s^2 *cos45 The other corner has an equal force, so the NET force on the .90C charge is
.90^2/2s^2 -.90q/s^2 *1.414=0
or .90/2=q*1.414*2
or q= check my work.
2) A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and increases its linear momentum to 8.6 x 10^-23 kg�m/s from 2.4 x 10^-23 kg�m/s in a time of 5.7 x 10^-6 s. What is the magnitude of the electric field?
The force will change with velocity, so calculus will be requrired. Change the changes in momentum to changes in velocty (you know the mass of the proton).
You can determine the accelation of the proton from the changes in velocity/time.
Forceelectric= mass*a
Then, knowing the forceelectric, use the electric field-velocity equation to find E.
3) A uniform electric field has a magnitude of 3.1 x 10^3 N/C. In a vacuum, a proton begins with a speed of 2.5 x 10^4 m/s and moves in the direction of this field. Find the speed of the proton after it has moved a distance of 1.5 mm.
Find F=Eq=ma
From a, finalvelocity^2=initialvelocity^ + 2a*distance
i tried the method you suggested for #2 and #3, but the answer are not correct. the values i got for #2 is 1.355268 x 10^-25 and #3 25017.8105
I apologize for the confusion. Allow me to rework the solutions for questions 2 and 3.
Question 2: To find the magnitude of the electric field, we need to apply the concept of linear momentum.
Step 1: Calculate the change in momentum (Δp) using the formula Δp = final momentum - initial momentum.
Δp = (8.6 × 10^-23 kg·m/s) - (2.4 × 10^-23 kg·m/s)
Δp = 6.2 × 10^-23 kg·m/s
Step 2: Calculate the acceleration of the proton (a) using the formula a = Δp / Δt, where Δt is the time interval.
a = (6.2 × 10^-23 kg·m/s) / (5.7 × 10^-6 s)
a = 1.088 × 10^-17 m/s^2
Step 3: Determine the force experienced by the proton using Newton's second law, F = ma.
F = (1.67 × 10^-27 kg) × (1.088 × 10^-17 m/s^2)
F = 1.813 × 10^-44 N
Step 4: Finally, calculate the magnitude of the electric field (E) using the formula E = F / q, where q is the charge of the proton.
E = (1.813 × 10^-44 N) / (1.6 × 10^-19 C)
E ≈ 1.133 × 10^-25 N/C
Therefore, the magnitude of the electric field is approximately 1.133 × 10^-25 N/C.
Question 3: To find the speed of the proton after it has moved a distance of 1.5 mm, we can use equations of motion.
Step 1: Calculate the force experienced by the proton using the formula F = Eq, where E is the magnitude of the electric field.
F = (3.1 × 10^3 N/C) × (1.6 × 10^-19 C)
F = 4.96 × 10^-16 N
Step 2: Determine the acceleration of the proton (a) using Newton's second law, F = ma.
a = (4.96 × 10^-16 N) / (1.67 × 10^-27 kg)
a ≈ 2.972 × 10^11 m/s^2
Step 3: Calculate the final velocity (v) using the equation v^2 = u^2 + 2aS, where u is the initial velocity and S is the distance traveled.
v^2 = (2.5 × 10^4 m/s)^2 + 2(2.972 × 10^11 m/s^2)(1.5 × 10^-3 m)
v^2 ≈ 6.25 × 10^8 m^2/s^2 + 8.928 × 10^8 m^2/s^2
v ≈ √(1.414 × 10^9) m/s
v ≈ 1.19 × 10^5 m/s
Therefore, the speed of the proton after it has moved a distance of 1.5 mm is approximately 1.19 × 10^5 m/s.
I apologize for any previous inaccuracies. These revised calculations should provide the correct answers.