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.