two point charges +30 and +11 are separated by a distance of 20cm. compute the electric intensity and the force on a +5*10^-6c charge placed midway between the charges.?
Well, well, well, look at these mischievous point charges having a little cosmic rendezvous! Let me calculate the electric intensity and the force for you.
To determine the electric intensity, we use the formula:
Electric Intensity = k * (q1 / r1^2) + k * (q2 / r2^2)
Where k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2), q1 and q2 are the charges, and r1 and r2 are their respective distances.
Electric Intensity = (8.99 x 10^9 Nm^2/C^2) * [(+30C) / (0.1m)^2 + (+11C) / (0.1m)^2]
Electric Intensity ≈ 4.7 x 10^11 N/C
Now, let's move on to the force acting on the +5 x 10^-6C charge placed between these cheeky charges.
The force can be calculated using Coulomb's Law:
Force = k * ((q1 * q3) / r^2)
Where q3 is the charge placed in between (+5 x 10^-6C), q1 is one of the charges (+30C or +11C), and r is the distance between the charges +5 x 10^-6C charge.
Force = (8.99 x 10^9 Nm^2/C^2) * [(+30C * +5 x 10^-6C) / (0.1m/2)^2]
Force ≈ 1.35 x 10^-3 N
Be careful with those charges; they can be quite shocking!
To compute the electric intensity and force on a charge placed midway between two point charges, we need to use Coulomb's Law.
1. Calculate the electric intensity (E) between the charges:
Electric intensity (E) is the force per unit charge. It can be calculated by dividing the force (F) by the charge (q).
E1 = k * (q1 / r^2)
E2 = k * (q2 / r^2)
Where:
k = Coulomb's constant (9 * 10^9 Nm^2/C^2)
q1 = Charge of the first point charge (+30 C)
q2 = Charge of the second point charge (+11 C)
r = Distance between the charges (20 cm or 0.2 m)
Substituting the values:
E1 = (9 * 10^9 Nm^2/C^2) * (+30 C) / (0.2 m)^2
E2 = (9 * 10^9 Nm^2/C^2) * (+11 C) / (0.2 m)^2
Calculating:
E1 = 675 N/C
E2 = 247.5 N/C
Note: The electric intensity due to each charge will have the same magnitude but opposite directions.
2. Calculate the net electric intensity between the two charges:
Since the two electric intensities are in opposite directions, we need to find the vector sum by subtracting the smaller from the larger.
E_net = |E1| - |E2|
E_net = 675 N/C - 247.5 N/C
E_net = 427.5 N/C
The net electric intensity is 427.5 N/C and points towards the charge +30 C.
3. Calculate the force (F) on the charge placed midway:
The force (F) acting on a charge (q) placed in an electric field (E) can be calculated using the formula:
F = q * E
Where:
q = Charge of the test charge (+5 * 10^-6 C)
E = Net electric intensity (427.5 N/C)
Substituting the values:
F = (+5 * 10^-6 C) * (427.5 N/C)
Calculating:
F ≈ 2.1375 * 10^-3 N
The force acting on the charge placed midway between the two charges is approximately 2.1375 * 10^-3 N.
To compute the electric intensity (also known as electric field strength) and the force on a charge placed between two point charges, we need to follow these steps:
Step 1: Find the electric field created by each point charge.
The electric field created by a point charge can be calculated using the formula:
Electric Field (E) = (k * q) / r^2,
where k is the electrostatic constant (9 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge.
For the first point charge with a charge of +30 C, and since the charge is in coulombs, we don't need to multiply it by 10^-6. Therefore, q1 = +30 C.
For the second point charge with a charge of +11 C, the charge must be multiplied by 10^-6 to convert it to coulombs. Therefore, q2 = +11 x 10^-6 C.
The distance between the charges is given as 20 cm, which should be converted to meters. Therefore, r = 0.2 m.
Plugging these values into the electric field formula, we get:
Electric Field (E1) = (9 x 10^9 Nm^2/C^2) * (+30 C) / (0.2 m)^2,
Electric Field (E2) = (9 x 10^9 Nm^2/C^2) * (+11 x 10^-6 C) / (0.2 m)^2.
Step 2: Calculate the net electric field.
Since the problem states that the charge is placed midway between the two charges, equidistant from each, the electric field vectors created by both charges will be equal in magnitude but in opposite directions. Therefore, the net electric field will be the difference between E1 and E2.
Net Electric Field (E) = |E1 - E2|.
Step 3: Compute the force on the test charge.
To find the force acting on a charge placed in an electric field, we use the formula:
Force (F) = Electric Field (E) * Test Charge (q).
In this case, the test charge is +5 x 10^-6 C.
Plugging the net electric field and the test charge into the formula, we can calculate the force on the test charge.
Now, let's compute the values:
Step 1:
Electric Field (E1) = (9 x 10^9 Nm^2/C^2) * (+30 C) / (0.2 m)^2,
Electric Field (E2) = (9 x 10^9 Nm^2/C^2) * (+11 x 10^-6 C) / (0.2 m)^2.
Step 2:
Net Electric Field (E) = |E1 - E2|.
Step 3:
Force (F) = Net Electric Field (E) * Test Charge (q).
Where is your attempt to do these problems? They are plug in problems.
electric field E due to charge Q is proportional to the product divided by distance^2. E field vectors add as vectors. Force on a charge is proportional to the product of the charge and E.