Two point charges lie on the x axis. A charge of 6.2μC is at the origin, and a charge of -9.5μC is at x=10.0cm. What is the net electric field at (a) x=−4.0cm and at (b) x=+4.0cm?
(a) At x=-4.0cm, the net electric field is -2.2 x 10^6 N/C.
(b) At x=+4.0cm, the net electric field is +2.2 x 10^6 N/C.
To find the net electric field at a point due to two point charges, we can use the principle of superposition. The net electric field at a point is the vector sum of the electric fields created by each individual charge at that point.
Given:
Charge at the origin (q1) = 6.2 μC
Charge at x=10.0cm (q2) = -9.5 μC
We can calculate the electric field at the specified points using the following formula:
Electric field (E) = k * q / r^2
Where:
k = Coulomb's constant = 9.0 x 10^9 Nm^2/C^2
q = charge
r = distance from the charge to the point
(a) x = -4.0cm
To calculate the net electric field at x = -4.0cm, we need to calculate the electric field due to each charge separately and then add them together.
For charge q1 at the origin:
r1 = distance from q1 to the point = |-4.0cm - 0| = 4.0cm = 0.04m (taking the absolute value to consider the direction)
E1 = (9.0 x 10^9 Nm^2/C^2) * (6.2 x 10^-6 C) / (0.04m)^2
For charge q2 at x = 10.0cm:
r2 = distance from q2 to the point = |10.0cm - (-4.0cm)| = 14.0cm = 0.14m
E2 = (9.0 x 10^9 Nm^2/C^2) * (-9.5 x 10^-6 C) / (0.14m)^2
The net electric field at x = -4.0cm is the vector sum of E1 and E2, E = E1 + E2.
(b) x = +4.0cm
To calculate the net electric field at x = +4.0cm, the process is the same as in part (a), but the distance to charge q2 changes.
For charge q1 at the origin:
r1 = distance from q1 to the point = |4.0cm - 0| = 4.0cm = 0.04m
E1 = (9.0 x 10^9 Nm^2/C^2) * (6.2 x 10^-6 C) / (0.04m)^2
For charge q2 at x = 10.0cm:
r2 = distance from q2 to the point = |10.0cm - 4.0cm| = 6.0cm = 0.06m
E2 = (9.0 x 10^9 Nm^2/C^2) * (-9.5 x 10^-6 C) / (0.06m)^2
The net electric field at x = +4.0cm is the vector sum of E1 and E2, E = E1 + E2.
Now let's calculate the values:
To calculate the net electric field at a given point due to multiple point charges, we use the principle of superposition. This principle states that the total electric field at a point is the vector sum of the electric fields created by each individual charge.
Let's break down the problem into two steps:
Step 1: Calculate the electric field at each point due to each individual charge.
Step 2: Add the electric fields to find the net electric field at each point.
Step 1: Electric Field Due to an Individual Charge
The electric field due to a point charge can be calculated using Coulomb's law:
E = k * (q / r^2)
where E represents the electric field, k is the electrostatic constant (k = 8.99 × 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge to the point where we want to calculate the electric field.
For the charge of 6.2 μC at the origin, the electric field at any point on the x-axis is:
E1 = k * (q1 / r1^2)
where q1 is +6.2 μC and r1 is the distance from the origin (0 cm) to the point on the x-axis.
For the charge of -9.5 μC at x = 10.0 cm, the electric field at any point on the x-axis is:
E2 = k * (q2 / r2^2)
where q2 is -9.5 μC and r2 is the distance from x = 10.0 cm to the point on the x-axis.
Step 2: Net Electric Field at a Point
To find the net electric field at a point, we need to consider the direction and magnitude of each electric field calculated in step 1.
(a) Net electric field at x = -4.0 cm:
To find the net electric field at this point, we need to consider the electric fields due to both charges.
The electric field due to the charge at the origin is E1, and the electric field due to the charge at x = 10.0 cm is E2.
Since the charges have the same sign (+6.2 μC and -9.5 μC), the electric fields will point in opposite directions.
The net electric field at x = -4.0 cm is the vector sum of E1 and E2:
E_net = E1 + E2
(b) Net electric field at x = +4.0 cm:
Again, to find the net electric field at this point, we need to consider the electric fields created by both charges.
The electric field due to the charge at the origin is E1, and the electric field due to the charge at x = 10.0 cm is E2.
Since the charges have opposite signs, their electric fields will point in the same direction.
The net electric field at x = +4.0 cm is given by the vector sum:
E_net = E1 + E2
By calculating the electric fields E1 and E2 at each point and adding them properly according to their directions, we can find the net electric field at x = -4.0 cm and x = +4.0 cm.