Answer: q = 2.781e-9C = 2.781nC
E=200C
Explanation:
E = Qd/(2πEor^3)
Where
E=Electric field intensity
Q=Charge
d=distance between the dipole=0.008m
Eo=permitivitty
400 N/C = Q(0.80e-2 m)/(2πε*(10e-2 m)^3)
Q= (400* 2* 3.142 * 8.85 x 10-12 * 0.1^3)/0.008
q = 2.781e-9C = 2.781nC
b)
Though the dipole are two separate charges. And since the point is on the x-axis, the electric field strengths are equivalent. The magnitude of the vector sum is:
E = kq*2sin θ/r^2
= 2(8.99e9 N*m^2/C^2)(2.781e-9 C)*sin(arctan(.4/10))/(10e-2 m)^2
= 2(8.99e9) * (2.781e-9) * sin(2.290)/(10e-2 m)^2
=200 C
(A) The value of charge is in nC is, [tex]2.84 \times 10^{-3} \;\rm nC[/tex].
(B) The magnitude of electric field at the point (0, 10) cm is 2.56 N/C.
Given data:
The distance between the two charges is, [tex]d = 0.800 \;\rm cm = 0.800 \times 10^{-2} \;\rm m[/tex].
The electric field strength at the point (0,10) cm is, E = 400 N/C.
(A)
In this part, we need to obtain the value of charge. To do so, we can use the formula of electric strength as,
[tex]E = \dfrac{kQ}{r^{2}}[/tex]
Here, k is the Coulomb's constant.
[tex]400 = \dfrac{9 \times 10^{9} \times q}{(0.800 \times 10^{-2})^{2}}\\\\q = 2.84 \times 10^{-12} \;\rm C\\\\q =2.84 \times 10^{-3} \times 10^{-3} \\\\q=2.84 \times 10^{-3} \;\rm nC[/tex]
Thus, the value of charge is in nC is, [tex]2.84 \times 10^{-3} \;\rm nC[/tex].
(B)
The electric field at the point (0, 10) cm is calculated as,
[tex]E = \dfrac{kq}{r'^{2}} \\\\E = \dfrac{9 \times 10^{9} \times 2.84 \times 10^{-12}}{(0-0.1)'^{2}} \\\\E = 2.56 \;\rm N/C[/tex]
Thus, the magnitude of electric field at the point (0, 10) cm is 2.56 N/C.
Learn more about the electric field strength here:
https://brainly.com/question/15170044
