Question

A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106 N/C. If the test charge is replaced with another test charge of 23 mC, what happens to the external electric field at P

Answer 1

The external electric field at point P does not change when the test charge is replaced.

Step-by-step explanation:

When the test charge is replaced with another test charge of 23 mC, the external electric field at point P does not change. The external electric field is determined by the source charges creating it and is independent of the test charge. Therefore, the magnitude and direction of the external electric field at point P remain the same.

When the test charge is replaced with another test charge of 23 mC, the external electric field at point P does not change. The external electric field is determined by the source charges creating it and is independent of the test charge. Therefore, the magnitude and direction of the external electric field at point P remain the same.

To calculate the total sound power P emitted by the source, we use the relationship between intensity I, power P, and the area A through which the power is spread. The formula for intensity is given by I = P/A, and for a spherical wave, the area A is a sphere with a surface area of 4πr², where r is the radius or distance from the source. Given the intensity I of 5.0 × 10⁻³ W/m² at a distance of 12 m, we can calculate the total power using:

P = I × (4πr²)

P = (5.0 × 10⁻³ W/m²) × (4π × (12 m)²)

P ≈ 7.2 W (rounded to one decimal place)

This means the total sound power emitted by the source is approximately 7.2 watts.

Answer 2

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Step-by-step explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

`E_2 = (E_1q_1)/(q_2)`

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

`E_2 = (4*10^6*13)/(23) = 2.26 *10^6 \ N/C`

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.