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In the figure below, four particles are fixed along an x axis, separated by distances d = 4.00 cm. The charges are q1 = +2e, q2 = -e, q3 = +e, and q4 = +5e, with e = 1.60 × 10-19 C. What is the value of the net electrostatic force on (a) particle 1 and (b) particle 2 due to the other particles?

q1 is at the origin and the rest of the particles are to the right a distance d from one another. therefore the order of particles is q1,q2,q3,q4 each with a distance d in between.
Got part a to be 5.596E-26 but i don't know how to do b. please help

  • physics-help - ,

    The only difference in part 2 from part 1 is that you have to consider space and direction of forces. Do this, sketch which way the forces are.

    at q2, reverse the sign of F since it is on the left side.

    Ftotal2= k/d^2 (-q1q2+q3q2+q4q2/2^2 )

    That should do it.

    If you want to the direction, You need to do the vector form of the equation.

  • physics-help - ,

    i ended up with 2.8767E-29 but its incorrect...i don't understand what im doing wrong still

  • physics-help - ,

    5. If the voltage between the two plates of an electric air cleaner is 500 V, how fast would a 10-12 kg soot particle with -1 „e 10-11 C of charge on it be moving if it went from the negative plate to the positive?

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