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 given point due to the two charges, we can use the formula for electric potential:
V = k * (q1/r1 + q2/r2)
Where:
V is the electric potential,
k is the electrostatic constant (8.99 * 10^9 Nm²/C²),
q1 and q2 are the charges,
r1 and r2 are the distances between the charges and the point at which we want to find the potential.
Let's calculate the potential at the given coordinates:
1. (0,0) cm:
The distance from both charges to this point is 10 cm. Thus, the potential at (0,0) cm is:
V = k * (q1/r1 + q2/r2)
= (8.99 * 10^9 Nm²/C²) * [(5 * 10^-6 C)/(10 * 10^-2 m) + (5 * 10^-6 C)/(10 * 10^-2 m)]
= 9.99 * 10^4 V
Therefore, the potential at (0,0) cm is 9.99 * 10^4 V.
2. (0,10) cm:
The distance from the charge at -10 cm to this point is 20 cm, and the distance from the charge at +10 cm is 30 cm. Thus, the potential at (0,10) cm is:
V = k * (q1/r1 + q2/r2)
= (8.99 * 10^9 Nm²/C²) * [(5 * 10^-6 C)/(20 * 10^-2 m) + (5 * 10^-6 C)/(30 * 10^-2 m)]
= 5.66 * 10^4 V
Therefore, the potential at (0,10) cm is 5.66 * 10^4 V.
3. (0,20) cm:
The distance from both charges to this point is 30 cm. Thus, the potential at (0,20) cm is:
V = k * (q1/r1 + q2/r2)
= (8.99 * 10^9 Nm²/C²) * [(5 * 10^-6 C)/(30 * 10^-2 m) + (5 * 10^-6 C)/(30 * 10^-2 m)]
= 3.77 * 10^4 V
Therefore, the potential at (0,20) cm is 3.77 * 10^4 V.
Now, let's consider the motion of a proton released from rest at the origin (0,0) cm. Protons are positively charged particles, so they will experience a repulsive force from the two positive charges. Since the potential is higher at (0,0) cm than at the other two points, the proton will experience a net force away from the origin. As a result, it will move away from the origin and accelerate in the positive direction.
To find the potential at a certain point due to two point charges, we can use the formula for the electric potential:
V = k * (q1 / r1 + q2 / r2)
where V is the potential, k is the electrostatic constant (8.99 x 10^9 N m^2/C^2), q1 and q2 are the charges, and r1 and r2 are the distances from the charges to the point where we want to find the potential.
In this case, we have two charges of 5 µC each, located at -10 cm and +10 cm on the horizontal axis. Let's calculate the potential at the given coordinates:
1. At (0,0) cm:
The distance from each charge to this point is 10 cm, so we have:
V1 = (8.99 x 10^9 N m^2/C^2) * (5 x 10^-6 C) / 0.1 m = 4.495 x 10^5 V
V2 = (8.99 x 10^9 N m^2/C^2) * (5 x 10^-6 C) / 0.1 m = 4.495 x 10^5 V
Since the charges are equal and opposite, the potential at this point is the sum of the potentials due to each charge:
V = V1 + V2 = 4.495 x 10^5 V + 4.495 x 10^5 V = 8.99 x 10^5 V
2. At (0,10) cm:
The distance from the charge at -10 cm is 10 cm, and the distance from the charge at +10 cm is 20 cm. Therefore:
V1 = (8.99 x 10^9 N m^2/C^2) * (5 x 10^-6 C) / 0.1 m = 4.495 x 10^5 V
V2 = (8.99 x 10^9 N m^2/C^2) * (5 x 10^-6 C) / 0.2 m = 2.2475 x 10^5 V
Again, since the charges are equal and opposite, the potential at this point is the sum of the potentials due to each charge:
V = V1 + V2 = 4.495 x 10^5 V + 2.2475 x 10^5 V = 6.7425 x 10^5 V
3. At (0,20) cm:
The distance from each charge to this point is 20 cm, so we have:
V1 = (8.99 x 10^9 N m^2/C^2) * (5 x 10^-6 C) / 0.2 m = 2.2475 x 10^5 V
V2 = (8.99 x 10^9 N m^2/C^2) * (5 x 10^-6 C) / 0.2 m = 2.2475 x 10^5 V
Again, since the charges are equal and opposite, the potential at this point is the sum of the potentials due to each charge:
V = V1 + V2 = 2.2475 x 10^5 V + 2.2475 x 10^5 V = 4.495 x 10^5 V
Now, regarding the proton released from rest at the origin, we know that protons are positively charged particles. In this scenario, the proton will experience a force due to the electric field created by the two equal and opposite charges. Since the potential decreases as we move farther away from the charges, the proton will be attracted towards the negative charge (at -10 cm) and repelled by the positive charge (at +10 cm). Therefore, the proton will move towards the -10 cm charge.