Two particles, one with negative charge -Q and the other with positive charge +2Q, are separated by a distance d. Both charges lie on the x-acis with the negative charge at x=0. At what point(s) along the line joining the particles is the potential equal to zero? At what point(s) along the line joining the particles is the electric field zero?
(a)
“Zero” point is between Q1 and Q2 and separated by distance ‘x’ from Q1.
φ=φ1+φ2= - k•Q1/x+k•Q2/(d-x) = 0,
k•Q1/x= k•Q2/(d-x)
Q/x=2Q/(d-x),
x=d/3.
(b) Assume that “zero” point is to the left from Q1 and separated by distance ‘x’ from Q1.
E1 is directed to the right, E2 is directed to the left
kQ1/ x²= kQ2/(d+x)²
Q/x² = 2Q/(d+x)²
x²-2dx- d²=0
x= d±√(d²+d²) =
= d±d√2.
x1= d(1+1.41) = 2.41d (this is the sought value)
x2 = d(1-1.41)= - 0.41d (extraneous root)
To find the point(s) along the line joining the particles where the potential is zero, we can use the equation for the electric potential due to a point charge:
V = kQ/r
Where V is the electric potential, k is the electrostatic constant (8.99 x 10^9 N⋅m^2/C^2), Q is the charge, and r is the distance from the charge.
Since we have two charges, we need to consider the potential due to each charge separately and then add them together.
For the negative charge -Q at x=0, the potential at any point x is:
V1 = k(-Q)/x
For the positive charge +2Q, the potential at any point x is:
V2 = k(2Q)/(d-x)
To find where the total potential is zero, we need to set V1 + V2 = 0 and solve for x:
(k(-Q)/x) + (k(2Q)/(d-x)) = 0
Now, we can solve this equation algebraically for x. Multiply both sides by x(d-x) to eliminate the denominators:
k(-Q)(d-x) + k(2Q)x = 0
Simplifying:
-Qd + Qx + 2Qx = 0
-Qd + 3Qx = 0
3Qx = Qd
x = d/3
So, along the line joining the particles, the potential is zero at x = d/3.
To find the point(s) along the line joining the particles where the electric field is zero, we can use the equation for the electric field due to a point charge:
E = kQ/r^2
Similar to finding the potential, we need to consider the electric field due to each charge separately and then combine them.
For the negative charge -Q at x=0, the electric field at any point x is:
E1 = k(-Q)/x^2
For the positive charge +2Q, the electric field at any point x is:
E2 = k(2Q)/(d-x)^2
To find where the total electric field is zero, we need to set E1 + E2 = 0 and solve for x:
(k(-Q)/x^2) + (k(2Q)/(d-x)^2) = 0
Multiply both sides by x^2(d-x)^2 to eliminate the denominators:
k(-Q)(d-x)^2 + k(2Q)x^2 = 0
Simplifying:
-Q(d-x)^2 + 2Qx^2 = 0
-Q(d^2 - 2dx + x^2) + 2Qx^2 = 0
-Qd^2 + 2Qdx - Qx^2 + 2Qx^2 = 0
- Qd^2 + 2Qdx + Qx^2 = 0
Qx^2 + 2Qdx - Qd^2 = 0
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
where a = Q, b = 2Qd, and c = -Qd^2.
Solving the equation:
x = (-2Qd ± √((2Qd)^2 - 4(Q)(-Qd^2)))/(2(Q))
x = (-2Qd ± √(4Q^2d^2 + 4Q^2d^2))/(2Q)
x = (-2Qd ± √(8Q^2d^2))/(2Q)
x = (-2d ± √(8d^2))(1/2)
Simplifying:
x = -d ± √2d
So, along the line joining the particles, the electric field is zero at x = -d ± √2d.