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Makovka662 [10]
1 year ago
6

What is the x-component of the total electric field due to these two charges at a point on the x-axis?

Physics
1 answer:
Y_Kistochka [10]1 year ago
4 0

Find the total x - component of the electric field at that point and get the resultant or the algebraic sum.

  • The electric field is a vector as it gives the force on a unit positive test charge placed at some distance from a charge.
  • The electric field due to each charge at that point on the x-axis can be found using the Coulombs formula: \vec{E} = \frac{1}{4\pi\epsilon}\frac{Q}{r^2}.
  • Then one needs to get the x-component of the electric field at the point due to each charge using the concept of the resolution of forces.
  • Once the x-components are found, one needs to get the resultant or the algebraic sum of them to calculate the x-component of the total electric field.

  • For extra knowledge, how to find the components of a given vector can be found here: brainly.com/question/28524840

#SPJ4

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A car with a mass of 2200 kg is travelling at a rate of 55 m's. What is the cars momentum? *
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\\ \sf\bull\dashrightarrow Momentum=Mass\times Velocity

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If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is t
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Answer:

The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Explanation:

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first magnetic field at first distance, B₁ = 2.50 mT

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Second magnetic field at Second distance, B₂ = ?

Second distance, r₂ = ?

Magnetic field for a straight wire is given as;

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Where:

μ is permeability

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r distance to the wire

Let \ \frac{\mu I}{2\pi}  \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

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