Point charges q1=50μCq1=50μC and q2=−25μCq2=−25μC are placed 1.0 m apart. (a) What is the electric field at a point midway between them? (b)

Question

Point charges q1=50μCq1=50μC and q2=−25μCq2=−25μC are placed 1.0 m apart. (a) What is the electric field at a point midway between them? (b) What is the force on a charge q3=20μCq3=20μC situated there?

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Thu Nguyệt 5 years 2021-08-17T07:39:54+00:00 1 Answers 242 views 0

Answers ( )

  1. Sapo1
    1
    2021-08-17T07:40:57+00:00
    Reply

    Answer:

    a) E = 2.7×10⁶ N/C

    b) F = 54 N

    Explanation:

    a) The electric field can be calculated as follows:

    E = \frac{Kq}{d^{2}}

    Where:

    K: is the Coulomb’s constant = 9×10⁹ N*m²/C²

    q: is the charge

    d: is the distance

    Now, we need to find the electric field due to charge 1:

    E_{1} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*50 \cdot 10^{-6} C}{(0.5 m)^{2}} = 1.8 \cdot 10^{6} N/C

    The electric field due to charge 2 is:

    E_{2} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*(-25) \cdot 10^{-6} C}{(0.5 m)^{2}} = -9.0 \cdot 10^{5} N/C

    The electric field at a point midway between them is given by the sum of E₁ and E₂ (they are in the same direction, that is to say, to the right side):

    E_{T} = E_{1} + E_{2} = 1.8 \cdot 10^{6} N/C + 9.0 \cdot 10^{5} N/C = 2.7 \cdot 10^{6} N/C to the right side                                                                                                

    Hence, the electric field at a point midway between them is 2.7×10⁶ N/C to the right side.  

    b) The force on a charge q₃ situated there is given by:

    E_{T} = \frac{F_{T}}{q_{3}} \rightarrow F_{T} = E_{T}*q_{3}

    F = 2.7 \cdot 10^{6} N/C*20 \cdot 10^{-6} C = 54 N

    Therefore, the force on a charge q₃ situated there is 54 N.  

    I hope it helps you!

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