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PLSSS PHYSICS HELP??? im past desperate!! Two charges, qA and qB, are at rest near a positive test charge, qT?

Two charges, qA and qB, are at rest near a positive test charge, qT, of 7.2 µC. The first charge, qA, is a positive charge of 3.6 µC, located 3.7 cm away from qT at 35°; qB is a negative charge of −5.5 µC, located 6.8 cm away at 139°. determine magnitude of force on qt and direction

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  • 9 years ago
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    Okay, haven't don't this for a bit... By the way, I hate this stuff. Here's my solution:

    Using Coulomb's Law.

    f_TA = 3.6u * 7.2u * K / (3.7e-2)^2 = 170.2nN

    f_TB = -5.5u * 7.2u * K / (6.8e-2)^2 = -76.97nN

    So we have:

    f_T = 170.2n /_ 35° - 76.97n /_ 139°

    That's the sum of the forces on f_T. If you're wondering what ' /_ ' is, it's my incredibly dodgy phasor symbol that I probably shouldn't be using because it implies that there's a complex part and the vector is spinning, which it isn't. I dunno what you're doing it, but we use it in Electrical Engineering for short hand.

    Anyways, now we convert the phasors to rectangular form, and add them.

    i and j are orthogonal unit vectors.

    f_T = 170.2n ( cos(35°) i + sin(35°) j) - 76.97n ( cos(139°) i + sin(139°) j )

    f_T = 197.5n i + 47.13 j

    Now we have the sum of the forces on f_T broken up into a vertical and horizontal component.

    For the magnitude:

    |f_T| = |197.5n i + 47.13 j| = 203.1nN

    For the direction:

    Arg(f_T) = Arg(197.5n i + 47.13 j) = 13.42°

    Well, there might be a mistake somewhere, but that's how you do it.

    By the way, what year is this for?

    Edit:

    Oops, left out the nano from K.

    Source(s): Electrical Engineering student from Australia.
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