In the picture above, the 3 charges Q1, Q2, and Q3 are located at positions (-a,
ID: 1426728 • Letter: I
Question
In the picture above, the 3 charges Q1, Q2, and Q3 are located at positions (-a,0), (a,0) and (0,-d) respectively. (The origin is the point halfway between Q1 and Q2.) Consider the special case where Q1, Q3 >0, and Q2=-Q1. Which of the following statements are true? The electric potential at the origin equals Q3/(4*?*?0*d). The electric field at the origin points in the positive y direction, away from Q3. The force on Q3 due to the other two charges is zero. The electric potential at any point along the y-axis is positive. The external work done to bring these charges to this configuration (from infinity) was positive. If Q3 is released from rest, it will initially accelerate to the right. The work required to move Q3 from its present position to the origin is zero
Explanation / Answer
-> Potential at a point due to a charge is given by (1/4o) (q/r)
since q1 and q3 are at equal distance from origin and their magnitude is equal but sign opposite, their potentials at origin will cancel (1/4o) (q1/a) +(1/4o) (-q1/a) = 0
Net resultant potential will be only due to q3 which is (1/4o) (q3/d). So, first statement true.
-> No, at the origin, the electric field due to q1 is in positive x direction and field due to q2 is in positive x direction. these two add up to give field in positive x direction. Next the field due to q3 is in positive y direction. So, adding this to the resultant field of the previous two charges the net field will be in the first quadrant,making some angle with the positive x axis.
-> The magnitudes of forces due to q1 and q2 on q3 will be equal but the directions will be along the line joining q1 and q3 (repulsive); and q3 and q2 (attractive) respectively. The vertical components of these forces cancel each other giving a resultant force in the positive x direction.
->Yes. Any point on the y axis is equidistant from the opposite charges q1 and q2 so the potential on any point on y axis due to these charges will be zero. Implying that, the net potential on any point on the y axis is only due to charge q3 which is positive regardless of the distance from the charge, since q3 is positive.
-> No. Work done to bring these charges to configuration is nothing but the potential energy of the system, which is given by
(1/4o) ( q1q2/r12 + q2q3 /r23 + q1q3 /r13 ) where, r12, r23, and r13 are the distances between charges q1 and q2; q2 and q3 ;and q1 and q3 respectively.
From the symmetry in the figure, r13 = r23
and q2 = -q1
=> PE = (1/4o) ( q1(-q1)/r12 + (-q1)q3 /r23 + q1q3 /r13 ) =
Since r13 = r23,
PE = (1/4o) ( -q12 /r12 ) = - (1/4o) ( q12 / 2a )
which is always negative. Hence the given statement is wrong.
-> Yes. As discussed earlier, the net force on q3 is in the positive x-direction. Hence intitally q3 will accelerate to the right is released.
-> Yes. If q3 is moved to origin, the final potential energy of the system will be, using the formula
(1/4o) ( q1q2/r12 + q2q3 /r23 + q1q3 /r13 ) ;
(1/4o) ( q1(-q1)/2a + (-q1)q3 /a + q1q3 /a ) = (1/4o) ( - q12 / 2a )
= - (1/4o) ( q12 / 2a ) which is the same as the PE of the system before moving q3. So, no work required to move the charge q3 to origin.
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