Two particles carrying 4c and -2c charges are placed on 1m long straight wire.determine the point on the line joining these particles where the electric potential is zero with reference to the positively charged particle.
kq1/r = kq2/1-r
k's cross. Solve for r
To determine the point on the line joining the particles where the electric potential is zero, we can use the principle of superposition. The electric potential at any point on the line is the sum of the electric potentials created by each of the charges individually.
Let's denote the distance from the positively charged particle as x.
The electric potential at a distance x from a point charge q is given by the equation V = k * q / r, where V is the electric potential, k is the electrostatic constant (9 × 10^9 Nm^2/C^2), q is the charge, and r is the distance.
For the positively charged particle with a charge of +4c, the electric potential is V1 = k * (4c) / x.
For the negatively charged particle with a charge of -2c, the electric potential is V2 = k * (-2c) / (1 - x).
To find the point where the electric potential is zero, we need the sum of these two potentials to be zero. Therefore,
k * (4c) / x + k * (-2c) / (1 - x) = 0.
Simplifying the equation, we have:
4 / x - 2 / (1 - x) = 0.
Now, we can solve this equation to find the value of x.
Multiply the entire equation by x(1 - x) to eliminate the denominators:
4(1 - x) - 2x = 0.
Expanding and simplifying the equation:
4 - 4x - 2x = 0.
-6x + 4 = 0.
Solving for x:
6x = 4,
x = 4 / 6,
x = 2 / 3.
Therefore, the point on the line where the electric potential is zero with reference to the positively charged particle is at a distance of 2/3 meters from the positively charged particle.