Respuesta :
Answer:
a)Q= + 0.71 nC , For the resultant electric field at the origin to be 45.0 N/C in the +x direction
b)Q= -2.83nC ,for the resultant electric field at the origin to be 45.0 N/C in the −x direction
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Equivalences
1nC= 10⁻9 C
Data
k = 8.99*10⁹ N×m²/C²
q₁ =+5.45nC = 3*10⁻⁹C
d₁ =1.35 m
d₂ = 0.595m
a)Problem development : sign and magnitude of Q for the resultant electric field at the origin to be 45.0 N/C in the +x direction
We make the algebraic sum of fields at at the origin :
[tex]E_{o} =E_{q} +E_{Q}[/tex] Equation (1)
[tex]E_{q} =\frac{k*q_{1} }{d_{1}{2} }[/tex]
Calculation of E(q)
[tex]E_{q} =\frac{8.99*10^{9} *5.45*10^{-9} }{1,35^{2} }[/tex]
[tex]E_{q} =26.88\frac{N}{C}[/tex] : in the +x direction .As the charge is negative, the field enters the charge
We replace [tex]E_{o}[/tex] and [tex]E_{q}[/tex] in the equation (1)
[tex]45=26.88+E_{Q}[/tex]
[tex]E_{Q} =45-26.88[/tex]
[tex]E_{Q} = 18.12 N/C[/tex] : in the +x direction .
Sign and magnitude of Q
Q must be positive for the field to abandon the load in the +x
[tex]E_{Q} =\frac{k*Q}{d_{2}^{2} }[/tex]
[tex]18.12=\frac{8.99*10^{9}*Q }{0.595^{2} }[/tex]
[tex]Q=\frac{18.12*0.595^{2} }{8.99*10^{9} }[/tex]
Q=0.71*10⁻⁹ C =0.71 nC
b)Sign and magnitude of Q for the resultant electric field at the origin to be 45.0 N/C in the −x direction
We make the algebraic sum of fields at at the origin :
[tex]E_{o} =E_{q} +E_{Q}[/tex]
[tex]-45=26.88+E_{Q}[/tex]
[tex]-71.88=E_{Q}[/tex]
[tex]71.88=\frac{8.99*10^{9} *Q}{0.595^{2} }[/tex]
Q= 2.83*10⁻⁹ C
Q= -2.83nC
Q must be negative for the field to enters the charge in the −x direction
The magnitude and sign of Q is given by the required magnitude and
sign of the charge at the origin due to the sum of the charges.
Responses:
- The sign and magnitude of Q when the charge is 45 N/C in the +x direction is, Q ≈ 2.83 nC
- The sign and magnitude of Q when the charge is 45 N/C n the -x direction is, Q ≈ -713.4 pC
How can the charge of the two particles at the origin be found?
The charge at the origin is given as follows;
When the charge at the origin is 45.0 N/C, we have;
[tex]45 = \mathbf{\dfrac{8.99 \times 10^{9} \times -5.45 \times ^{-9} }{1.35^2} + \dfrac{8.99 \times 10^{9} \times Q} {(-0.595)^2}}[/tex]
Which gives;
[tex]Q = \dfrac{\left(45 - \dfrac{8.99 \times 10^{9} \times -5.45 \times ^{-9} }{1.35^2} \right) \times (-0.595)^2}{8.99 \times 10^{9} } \approx \mathbf{2.83 \times 10^{-9}}[/tex]
When the charge at the origin is [tex]E_0[/tex] = 45 N/C, we have;
- Q ≈ 2.83 × 10⁻⁹ C = 2.83 nC
When the charge at the origin is [tex]E_0[/tex] = 45 N/C in the -x direction, we have;
[tex]Q = \dfrac{\left(-45 - \dfrac{8.99 \times 10^{9} \times -5.45 \times ^{-9} }{1.35^2} \right) \times (-0.595)^2}{8.99 \times 10^{9} } \approx -7.134 \times 10^{-10}[/tex]
Therefore;
The charge at the origin is [tex]E_0[/tex] = 45 N/C in the -x direction, we have;
- Q ≈ -7.134 × 10⁻¹⁰ C = -713.4 pC
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