Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.235 m to the right of Q1. Q3 is located 0.125 m to the right of Q2. The force on Q2 due to its interaction with Q3 is directed to the.....

Right if the two charges are negative. A: True B: False

Right if the two charges have opposite signs. A: True B: False

Left if the two charges have opposite signs. A: True B: False

Left if the two charges are positive. A: True B: False

Left if the two charges are negative. A: True B: False

Part 2

In the above problem, Q1= 1.22·10-6 C, Q2= -3.03·10-6 C, and Q3= 3.33·10-6 C. Calculate the total force on Q2. Give with the plus sign for a force directed to the right. (in N)

Part 3

Now the charges Q1= 1.22·10-6 C and Q2= -3.03·10-6 C are fixed at their positions, distance 0.235 m apart, and the charge Q3= 3.33·10-6 C is moved along the straight line. For what position of Q3 relative to Q1 is the net force on Q3 due to Q1 and Q2 zero? Use the plus sign for Q3 to the right of Q1.
(in m)

Please solve and explain problem.

Solution Summary

Three questions are involved in this exercise.
Question 1) three charges are placed in a straight line. The directions of the net force on each charge are explored.
Question 2) three charges are placed in a straight line. The net force on one of the point charge is calculated.
Question 3) three charges are placed in a straight line. Find the location of a point charge such that its net force is zero.

... the GDP - Compute the net exports 3) Using the ... Graphically, this is expressed at the point of intersection ... a. An increase in Japanese interest rates b. An ...

... Give electric fields in N/C and charges in Coulombs. ...3. A small, non-conducting ball of mass m = 1.3 mg and ... mg Electric field at any point due to an infinite ...

... In (3) all terms on the RHS are known, hence ... conducting sphere is due to induced charges because of ... The force between a point charge and conducting sphere is ...

... b2 - 2abcosθ)3/2][(a2 - b2)/a] a2sinθ dθ Π = [a(a2 - b2)Q/4ΠЄ0] ∫sinθ/(a2 + b2 - 2abcosθ)3/2] dθ ... Potential at point P due to two charges Q and Q ...

... kq (r - r¢) (1.1) E= 3 r - r ... If the charge is positive, the field points away from the charge. In our case we have two equal (negative) charges located in this ...

... Rates of deceleration are always expressed as _____. ... far is the train located from its starting point. ... There are three forces on the truck: gravity mg ...

...3. A circular loop, 10.0 cm in diameter, rotates from ... As always, assume the currents are positive charges. ... magnetic field produced by the wires at point P, 4.0 ...

... The table gives the masses, charges and speeds of the 11 particles that take these paths through the field in the direction ... I1 Solution 3: ... (c) at point P, 5 cm ...

... As the firm tries to attract more new dollars, the cost of each dollar will at some point rise. ... Straight line depreciation rates for 5 years. Year 1 2 3 4 5. ...