Open Question: Two equal point charges of 5 µC are on the horizontal axis. One charge is at -10 cm, the other is at +10 cm.?
These two charges don't move, they are STUCK, take the potential to be zero at infinity.
Find the potential at the following coordinates: (0,0) cm, (0,10)cm & (0,20)cm.
What will happen to a proton released from rest at the origin?
To find the potential at a specific point due to two point charges, we can use the formula for electric potential:
V = k * (q1 / r1 + q2 / r2)
Where:
V is the potential at the point,
k is the electrostatic constant (9 × 10^9 N m^2/C^2),
q1 and q2 are the charges,
r1 and r2 are the distances from the charges to the point.
Let's calculate the potential at each of the given coordinates.
1. Potential at (0,0) cm:
Since the point is at the origin, the distances from the charges to this point are both 10 cm. The charges are equal, so we can substitute their values into the formula:
V = (9 × 10^9 N m^2/C^2) * (5 × 10^(-6) C / 10 cm + 5 × 10^(-6) C / 10 cm)
= (9 × 10^9 N m^2/C^2) * (1 × 10^(-6) C / 0.1 m)
= 90 V
Therefore, the potential at (0,0) cm is 90 volts.
2. Potential at (0,10) cm:
In this case, the distance from the first charge is 20 cm, and the distance from the second charge is 10 cm. Using the formula:
V = (9 × 10^9 N m^2/C^2) * (5 × 10^(-6) C / 0.2 m + 5 × 10^(-6) C / 0.1 m)
= (9 × 10^9 N m^2/C^2) * (1 × 10^(-6) C / 0.2 m)
= 45 V
Therefore, the potential at (0,10) cm is 45 volts.
3. Potential at (0,20) cm:
Here, the distance from the first charge is 30 cm, and the distance from the second charge is 20 cm. Plugging the values into the formula:
V = (9 × 10^9 N m^2/C^2) * (5 × 10^(-6) C / 0.3 m + 5 × 10^(-6) C / 0.2 m)
= (9 × 10^9 N m^2/C^2) * (1 × 10^(-6) C / 0.3 m)
= 30 V
As a result, the potential at (0,20) cm is 30 volts.
Now, let's discuss what will happen to a proton released from rest at the origin. Since the origin has a positive potential (90 volts), the proton will move away from the origin towards lower potential regions. Therefore, the proton will move towards either (0,10) cm or (0,20) cm, where the potential is 45 volts and 30 volts, respectively.