Respuesta :
Answer:
r = 8/2 = 4cm = 0.04m
k = 9×10^9
Enet = 51 N/C
Enet = E1 + E2
since E1 = E2
E1 = Enet/2 = 51/2
E/2 = kq/r²
q = Er²/2k
q = (51 × 0.04²)/(2×9×10^9)
q = 4.5×10^-12 C
q1 = q2 = 4.5 pC
Explanation:
The electric field is a region around a
charge in which it exerts electrostatic force
on another charges. While the strength of
electric field at any point in space is called
electric field intensity. It is a vector
quantity. Its unit is NC¯¹.
According to coulomb’s law ,if a unit
positive charge q (call it a test charge) is
brought near a charge q (call a field
charge) placed in space,the charge q will
experience a force. The value of this force
depends upon the distance between the
two charges. If the charge q is moved
away from q ,this force would decrease till
at a certain distance the force would be
practically reduced to zero. The charge q
is then out of the influence of charge q.
The region of space surrounding the charge
q in which it exerts a force on the charge
q is known as E.F of the charge
q. Mathematically it is expressed as:
E =F/q
The direction of the vector E is the same
as the direction of F,because q is a
positive scalar. Dimensionally,the E.F is
force per unit charge,and its SI unit is the
newton/coulomb (N/C).
Electric charge is a basic property of electrons, protons, and other subatomic particles. The magnitude of point charges are [tex]Q_1 = Q_2 = 1.81\times 10^{-11}\;\rm C[/tex].
What is the charge?
The charge is a characteristic of a unit of matter that expresses the extent to which it has more or fewer electrons than protons.
Given that the magnitude of the combined electric field E is 51 N/C. Also, Two point charges of equal magnitude are 8.0 cm apart.
[tex]E = E_1 + E_2[/tex]
The magnitude of both the charges are the same, hence the magnitude of the electric field around the charges will be the same.
[tex]E_1 = E_2[/tex]
So, [tex]E = 2E1[/tex]
[tex]E_1 = \dfrac {E}{2}[/tex]
[tex]E_1 = E_2 = \dfrac {51}{2}[/tex]
[tex]E_1 = E_2 = 25.5\;\rm N/C[/tex]
The electric field is around the charge is given as,
[tex]E = \dfrac {kQ}{d^2}[/tex]
Where Q is the magnitude of the point charge, d is the distance between the charge and k is the coulombs' constant.
[tex]25.5 = \dfrac {9\times 10^9\times Q}{0.08^2}[/tex]
[tex]Q = 1.81 \times 10^{-11} \;\rm C[/tex]
The point charges are equal in magnitude, hence,
[tex]Q_1 = Q_2 = 1.81\times 10^{-11}\;\rm C[/tex]
Hence we can conclude that the magnitude of point charges are [tex]Q_1 = Q_2 = 1.81\times 10^{-11}\;\rm C[/tex].
To know more about the charges, follow the link given below.
https://brainly.com/question/24391667.
