Answer:
q = 3.703*10^-9
dipole moment = 3.703*10^-11
E @ (10,0) = 331.977 N/C
Explanation:
Given:
Coordinates
+ q = (0,-0.5) cm
- q = (0,0.5) cm
E = 360 N/C @ (0,10)
Solution:
[tex]Electric Field Strength @ (0,10) = \frac{k*q}{R^2}\\[/tex]
Note: +q (away from charge) and -q (towards the charge)
[tex]E = k*q*(\frac{1}{R^2_{1}} - \frac{1}{R^2_{2} })\\ 360 = (9*10^9)*q*(\frac{1}{0.095^2} - \frac{1}{0.105^2 })\\\\q =1.99 *10^-9[/tex]
Dipole moment = q.(space between two charges)
= (3.703*10^-9 * 0.01)
= 3.703*10^-11
Electric Field Strength @ (10,0)
[tex]E = 2*k*q*(\frac{1}{R^2})*cos(Q)\\ \\Q = arctan(\frac{10}{0.5}) = 87.138 degrees\\R = \sqrt{10^2 + 0.5^2} = 10.01249 cm\\E = 2* (9*10^9)*(3.703*10^-9)*\frac{cos (87.138)}{0.1001249^2}\\E = 331.977 N/C[/tex]
Answer:
a. dipole moment = 2.003 × 10⁻¹³ Cm, charge = 2.003 × 10⁻¹¹ C b. 8.96 N/C
Explanation:
a. The electric field due to a dipole is given by
E = p/2πεy³ where p = dipole moment and y = distance of dipole to point of electric field = 10 cm = 0.10 m. E = electric field strength at (0,10) = 360 N/C
So, E = p/2πεy³
p = 2πεy³E = 2π × 8.854 × 10⁻¹² × (0.10 m)³ × 360 N/C = 2.003 × 10⁻¹³ Cm
Also, p = qd where q = charge and d = distance of charges apart = 1.0 cm = 0.01 m
q = p/d = 2.003 × 10⁻¹³ Cm/0.01 m = 2.003 × 10⁻¹¹ C
b. The electric field at point (10,0)
E = qx/[πε√(d² + 4x²)]³ which is the electric field at an axis perpendicular to the dipole where x = 10 cm = 0.10 m
E = 2.003 × 10⁻¹¹ C × 0.10/π × 8.854 × 10⁻¹²√(0.01² + 4×(0.1)²)³ = 8.96 N/C.
