The following questions use a central charge of q1=+2x10^-4C
1. What is the field strength at a distance of 5m from the central charge?
I got 7.2x10^4 However, I'm not sure which unit to use... is it joules?
2. What is the force acting on another charge (+5x10^-4C) which is 5m from the central charge?
I'm not sure which formula to use. Would it be Fe=(kq1q2)/r^2 ? It just doesn't seem right; would I need to use vectors to find the direction?
3. What is the electrical potential for the charge in question #2 if it is moved from infinity to a distance 3m from the central angle?
This confuses me because I thought it was 5m from the central charge, or is this different?
4. What is the difference in electrical potential if the charge from question #2 is moved from 10m away from the central charge to 2m away from it?
1. If you used E = k q/R^2, the field strength should be in Newtons per Coulomb or Volts per meter, which are the same thing
2. That is the correct formula
3. It's a different question, and location. The potential (in volts) is
k q1/ R , regardless of the value of q2
4. Compute the potential difference between the two locations
It is not clear if, when they are asking for electric potential (J/C), they really want potential energy (J)
To answer these questions, you need to apply the concepts of electrostatics and use relevant formulas. Let's break it down one question at a time:
1. To find the field strength at a distance of 5m from the central charge, you can use the formula for electric field (E) which is given by E = k * (q1/r^2), where k is the electrostatic constant (8.99x10^9 Nm^2/C^2), q1 is the central charge, and r is the distance from the charge. Plugging in the values, we get E = (8.99x10^9 Nm^2/C^2) * (2x10^-4C / (5m)^2). By performing the calculations, you find E ≈ 7.2x10^4 N/C. The unit for electric field strength is Newtons per Coulomb (N/C), not Joules.
2. To find the force acting on another charge (+5x10^-4C) that is 5m from the central charge, you can use Coulomb's law. The formula is F = (k * q1 * q2) / r^2, where F is the force, q2 is the charge experiencing the force (in this case, +5x10^-4C), and r is the distance between the charges. Plugging in the values, we get F = (8.99x10^9 Nm^2/C^2) * (2x10^-4C) * (5x10^-4C) / (5m)^2. After calculating, you will find F ≈ 1.798 N. It is important to note that force is a vector quantity, so you should consider its direction as well.
3. The question asks for the electrical potential for the charge from question #2 when it is moved from infinity to a distance of 3m from the central charge. The formula for electrical potential (V) due to a point charge is given by V = k * (q1 / r), where V is the potential, q1 is the central charge, r is the distance from the charge, and k is the electrostatic constant. Plugging in the values, we get V = (8.99x10^9 Nm^2/C^2) * (2x10^-4C / 3m). After performing the calculations, you will find V ≈ 5.992x10^4 V.
4. The difference in electrical potential when the charge from question #2 is moved from 10m to 2m away from the central charge can be calculated using the same formula as question #3. You need to find the potential at each distance and then subtract the initial potential from the final potential. The initial potential is V1 = (8.99x10^9 Nm^2/C^2) * (2x10^-4C / 10m), and the final potential is V2 = (8.99x10^9 Nm^2/C^2) * (2x10^-4C / 2m). Subtracting V1 from V2 will give you the difference in electrical potential.