Question

An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented along the y-axis. What is the electric field strength at point (x,y) = ( 20 mm ,0cm)? What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer 1

The electric field strength at point (x,y) = ( 20 mm ,0cm) due to an electric dipole is 11.25 N/C. The electric field strength at point (x,y) = (0cm, 20 mm) is also 11.25 N/C.

Step-by-step explanation:

The electric field strength at point (x,y) = ( 20 mm ,0cm) due to an electric dipole can be calculated using the formula for the electric field of a point charge. In this case, we have two charges of ± 5.0 nC separated by 3.0 mm. The electric field at the given point can be found using the formula:

E = (k * q) / r^2

Here, k is the electrostatic constant, q is the charge, and r is the distance from the charge. Plugging in the values, we get:

E = (9 * 10^9 N*m^2/C^2) * (5.0 * 10^-9 C) / (0.02 m)^2

E = 11.25 N/C

Therefore, the electric field strength at point (x,y) = ( 20 mm ,0cm) is 11.25 N/C.

The electric field strength at point (x,y) = (0cm, 20 mm) will be the same as the field at (20 mm, 0cm) since the dipole is oriented along the y-axis and the field is symmetrical along that axis. Therefore, the electric field strength at point (x,y) = (0cm, 20 mm) is also 11.25 N/C.

Answer 2

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =16321.0769 N/C

The electric field strength at point (x,y) = (0cm, 20 mm) is =35321.58999 N/C

Step-by-step explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=16321.0769 N/C

Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷ 0.000000152881

=35321.58999 N/C