Charges of - q and +2 q are fixed in place, with a distance d between them (see
ID: 2245589 • Letter: C
Question
Charges of -q and +2q are fixed in place, with a distance d between them (see the drawing). A dashed line is drawn through the negative charge, perpendicular to the line between the charges. On the dashed line, at a distance L from the negative charge, there is at least one spot where the total potential is zero.
Charges of -q and +2q are fixed in place, with a distance d = 2.65 m between them. A dashed line is drawn through the negative charge, perpendicular to the line between the charges. On the dashed line, at a distance L from the negative charge, there is a spot where the total potential is zero.
2. Find the distance L.
Number Unit L = ---Select---mm^2m^3 Charges of -q and +2q are fixed in place, with a distance d between them (see the drawing). A dashed line is drawn through the negative charge, perpendicular to the line between the charges. On the dashed line, at a distance L from the negative charge, there is at least one spot where the total potential is zero. Charges of -q and +2q are fixed in place, with a distance d = 2.65 m between them. A dashed line is drawn through the negative charge, perpendicular to the line between the charges. On the dashed line, at a distance L from the negative charge, there is a spot where the total potential is zero. What is the algebraic expression for the total potential VTotal at a given spot (not necessarily the spot where the potential is zero) that is produced by two point charges q1 and q2? Express your answer in terms of q2 and q2, the distance r1 from the charge q1 to the spot, the distance r2 from the charge q2 to the spot, and the constant k. (Answer using q_1 for q1, q_2 for q2, r_1 for r1, r_2 for r2, and k.) Find the distance L.
Explanation / Answer
Solution to Q1:
Since, potential due to one charge is independent of the location of other charge. We can treat each of the charges independently. This is called Super Position Principle.
Potential due to charge 1(V1)=(q_1)/4*pi*e*r_1
Here q_1 is charge, r_1 is the distance of the point from the charge, e is the permittivity of the medium.Similar thing can be done for charge q_2.
Therefore, potential due to the combination of charges is
V= V1 + V2.
=> V= (q_1)/4*pi*e*r_1 + (q_2)/4*pi*e*r_2
Solution to Q2:
Equating the above calculated potential to zero for
q_1= -q, r_1= L, q_2= 2*q, r_2=sqrt(L^2 + d^2)
we get
2*L=sqrt(L^2 + d^2)
=> L = +d/sqrt(3) or -d/sqrt(3).
Substituting d = 2.65 cm.
we get
L = 1.53cm perpendicularly above or perpendicularly below.