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Two charges, qA and qB, are at rest near a positive test charge, qT, of 7.2 µC. The first charge, qA, is a pos?
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
1 Answer
- bpiguyLv 79 years agoFavorite Answer
The force on qT due to qA is k qA qT / a^2 = k (3.6) (7.2) / 3.7^2 at an angle of 215°.
The force on qT due to qB is k qB qT / b^2 = k (5.5) (7.2) / 6.8^2 at an angle of 139°.
Combining these as vectors, the force is
7.2k <(-3.6/3.7^2 sin 35° + 5.5/6.8^2 sin 41°) i - (3.6/3.7^2 cos 35° + 5.5/6.8^2 cos 41) j>
= <(-1.0860 + 0.5618) i - (1.5509 + 2.1973) j> = <-0.5241 i - 3.7482 j> = 3.7847 k (partial answer)
The angle is arctan (0.5241 / 3.7482) =188° (partial answer -- third quadrant)
Now the units have to be accounted for. The constant k = 9.0 x 10^9 N m2/C2. The distances in cm have to be converted to meters (x 10^-2), and the charges in μC have to be converted to C (x 10^-6).
After you've done that, your answer will be in newtons.
