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Sagot :
Answer:
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
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
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