Joe asked in Science & MathematicsPhysics · 1 month ago

Physics Help?!?

A 7.00 microcoulomb charge is .500 m above a 15.0 microcoulomb charge. Calculate the electric field at a point 1.00m above the 15.0 microcoulomb charge.

Charges of -3.50 nC and -7.00 nC are 20.0cm apart. Find a 5.0 nC charge’s equilibrium position.

1 Answer

  • 1 month ago

    E fields add as vectors

    assuming both charges are +

    from the first charge

    E = k7µ/0.5² = 28e-6(9e9) V/m pointed up

    E = 252 kV/m pointed up

    from the second chargeE = k15µ/1² = 15e-6(9e9) V/m pointed upE = 135 kV/m pointed up

    sum = 387 kV/m pointed up

    Electric field

    The strength or magnitude of the field at a given point

    is defined as the force that would be exerted on apositive test charge of 1 coulomb placed at that point;

    the direction of the field is given by the direction ofthat force.

       E = F/Q = kQ/r²

       Q = F/E   F = QE

    in Newtons/coulomb OR volts/meter

       k = 1/4πε₀ = 8.99e9 Nm²/C²

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