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A charge q1 of +3.0 x10^-6C is held at (-12, 0) (cm) . Another charge q2...

A charge q1 of +3.0 x10^-6C is held at (-12, 0) (cm) . Another charge q2 of the same magnitude is held at (12,0). The third charge q3 is held at (0,8) = -6x10^16C

a) What is the magnitude of the net force on charge q1 and q2

b) If q3 were to be released what direction and angle will it move in

c) At what point on the plane would a charge q4 of +15x 10^-6 be placed so that net force is 0

Solutions

Expert Solution

You have mentioned . But i think this is . If i am wrong please comment.

The electric force between two charged particles having charge ' q ' and ' Q ' and separated by a distance ' r ' is given by

  

Where,

Hence, if we release the charge then it will move in negative y - direction . i.e. making an angle of with positive x - axis in anticlockwise direction.

(c ) Please mention, on which charge the net force is zero and re upload the question.

genius_generous answered 5 months ago
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