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?
The potential at any point is the sum of the potentials due to the two separate charges. (that is the thing about potentials, they add)
due to charge 1
k = 1/(4 pi eo)
Q = 5 u C
U1 = k Q /((x+.1)^2+y^2)^.5
due to charge 2
U2 = k Q /[( x-.1)^2 +y^2) ]^.5
so
at (0,0)
U = U1 + U2 = k Q [1/.1^2+1/.1)^2
at (0,10)
U = U1+U2= k Q [1/(.1^2+.1^2)^.5 + same]
at (0,20)
k Q [ 1/(.1^2+.2^2)^.5 + same]
What do u mean by + same
To find the potential at a given coordinate, we need to use the formula for the electric potential due to a point charge. The formula is:
V = k * q / r
Where:
V is the electric potential,
k is the Coulomb's constant (k = 9 * 10^9 N m^2 / C^2),
q is the charge, and
r is the distance from the charge to the coordinate point.
Let's calculate the potentials at the given coordinates:
1. For the point (0,0) cm:
The distance from the point charge at -10 cm is 10 cm, and the distance from the point charge at +10 cm is also 10 cm (since they are equidistant from the origin). The charges are equal and opposite, so their potentials will cancel each other out. Therefore, the net potential at (0,0) cm is zero.
2. For the point (0,10) cm:
The distance from the point charge at -10 cm is 20 cm, and the distance from the point charge at +10 cm is 10 cm. Using the formula, we can calculate the potential due to each charge and add them together:
V_net = (k * q) / r1 + (k * q) / r2
V_net = (9 * 10^9 N m^2 / C^2) * (5 * 10^(-6) C) * (1 / 0.2 m) + (9 * 10^9 N m^2 / C^2) * (5 * 10^(-6) C) * (1 / 0.1 m)
V_net = 9 * 10^8 N m / C + 1.8 * 10^9 N m / C
V_net = 2.7 * 10^9 N m / C
So, the potential at (0,10) cm is 2.7 * 10^9 V.
3. For the point (0,20) cm:
The distance from the point charge at -10 cm is 30 cm, and the distance from the point charge at +10 cm is 20 cm. Using the same formula as above, we can calculate the potential due to each charge and add them together:
V_net = (k * q) / r1 + (k * q) / r2
V_net = (9 * 10^9 N m^2 / C^2) * (5 * 10^(-6) C) * (1 / 0.3 m) + (9 * 10^9 N m^2 / C^2) * (5 * 10^(-6) C) * (1 / 0.2 m)
V_net = 3 * 10^8 N m / C + 2.25 * 10^9 N m / C
V_net = 2.55 * 10^9 N m / C
So, the potential at (0,20) cm is 2.55 * 10^9 V.
Now, let's analyze what will happen to a proton released from rest at the origin (0,0) cm. A proton has a positive charge. Since the potentials at (0,0) cm and (0,20) cm are both positive, and the potential increases as we move away from the charges, the proton will experience a repulsive force from the charges and will be pushed away. Therefore, the proton will move away from the origin along the positive y-axis.