The middle two question please check my answers and tell me if I am correct or if there is something I am missing. The first and the last question I need help to understand and complete it. Thank you in advance!!!
1. What is the electric force between a +2 µC point charge and a –2 µC point charge if they are separated by a distance of 5.0 cm? Show your work. (µC = 1.0 × 10–6 C)
Answer:
2. The strength of an electric field at 5.0 cm from a point charge is 100.0 N/C. What is the magnitude of the source charge? Show your work.
Answer:
5 cm. r = 5/100 = 0.05 m
E=q4πϵ∗r2
Constant: 9 * 10^9
100=9∗109∗q0.05
q=0.55∗10−9C
3. A 5.0 µC point charge is moved within an electric field and has an electric potential energy change of 10.0 J. What is the electric potential difference before and after the charge was moved? Show your work. (µC = 1.0 × 10–6 C)
Answer:
10 = [1.0 • 10^ (- 6) ] • (V)
V = 1.0 • 10^7 volts
Delta v 10.0 µJ
4. Complete each of the following parts and show your work:
a.) Draw the electric field around a +2.0 × 10–5 C point source charge. Draw the field lines and indicate their direction. Draw two different equipotential lines.
b.) Calculate the force between a +5 µC test point charge and this source charge at a distance of 2.00 cm. (µC = 1.0 × 10–6 C)
c.) If the test charge were moved closer to the source charge, would the change in its potential energy be positive, negative or zero? Explain.
Answer:
1. To calculate the electric force between two point charges, you can use Coulomb's Law equation:
F = (k * |Q1| * |Q2|) / r^2
Where:
- F is the electric force
- k is Coulomb's constant (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
In this case, the magnitudes of the charges are +2 µC and -2 µC, which correspond to 2 * 10^-6 C and -2 * 10^-6 C respectively. The distance between them is 5.0 cm, which is equivalent to 0.05 m.
Substituting these values into the equation:
F = (9 * 10^9 * |2 * 10^-6| * |-2 * 10^-6|) / (0.05)^2
F = (9 * 10^9 * 2 * 10^-6 * 2 * 10^-6) / 0.0025
F = 7.2 * 10^-2 N
So, the electric force between the two charges is 7.2 * 10^-2 N.
2. To calculate the magnitude of the source charge when given the strength of an electric field at a distance, you can rearrange the equation for electric field:
E = (k * |Q|) / r^2
- E is the electric field strength
- k is Coulomb's constant (9 * 10^9 N m^2/C^2)
- |Q| is the magnitude of the charge
- r is the distance from the charge
In this case, the electric field strength is 100.0 N/C and the distance is 5.0 cm, which is equivalent to 0.05 m. We need to solve for |Q|.
Substituting the values into the equation:
100.0 = (9 * 10^9 * |Q|) / (0.05)^2
100.0 = (9 * 10^9) * |Q| / 0.0025
|Q| = (100.0 * 0.0025) / (9 * 10^9)
|Q| = 2.78 * 10^-12 C
So, the magnitude of the source charge is 2.78 * 10^-12 C.