The force exerted by q1 on q3 is vector F 13. The force exerted by q2 on q3 is v

Question

The force exerted by q1 on q3 is vector F 13. The force exerted by q2 on q3 is vector F 23. The resultant force vector F 3 exerted on q3 is the vector sum vector F 13 + vector F 23.
Goal Apply Coulomb's law in two dimensions.

Strategy Coulomb's law gives the magnitude of each force, which can be split with right-triangle trigonometry into x- and y-components. Sum the vectors componentwise and then find the magnitude and direction of the resultant vector.

LEARN MORE
Remarks The methods used here are just like those used with Newton's law of gravity in two dimensions.

Question Without actually calculating the electric force on q2, determine the quadrant into which the electric force vector on q2 points.
quadrant 1 where 0 ? ? < 90? quadrant 2 where 90? ? ? < 180? quadrant 3 where 180? ? ? < 270? quadrant 4 where 270? ? ? < 360? Your answer is correct.

PRACTICE IT
Use the worked example above to help you solve this problem. Consider three point charges at the corners of a triangle, as shown in the figure, where q1 = 5.53 10-9 C, q2 = -2.43 10-9 C, and q3 = 5.25 10-8 C.

EXERCISE
Using the same triangle, find the vector components of the electric force on q1 and the vector's magnitude and direction. (Use the charges given in the Practice It section.)
Fx = .....????
Fy = ?????
Go back to your diagram of the system. Use the directions of the force arrows help you to estimate the direction and relative size of the total force. Then check the vector addition. N
magnitude ???? N
direction ?????? counter-clockwise from the +x-axis

Explanation / Answer

force on q1 charge F12 = force on 1 by 2 = 13.4379 * (10^-9) j F13= force on 1 by 3 = - 80.224 *(10^-9) i - 62.899 *(10^-9) j net force on q1 = F1 = - 80.224 *(10^-9) i -49.4611 *(10^-9) j N Fx = - 80.224 *(10^-9) Fy = -49.4611 *(10^-9) magnitude = 94.245 *(10^-9) N

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