Two point charges of +40x10^-6C and -60x10^-6 are placed 50 cm apart. What is the distance from +40x10^-6 charge along the line joining the two point charges if the resultant potential is zero..
Can you give me the exact formula to do this question and explain more on it.. Thank you..
potentials add
Q1/r = Q2/(.5-r)
To find the distance from the +40x10^-6C charge along the line joining the two charges where the resultant potential is zero, we can use the formula for electric potential due to a point charge at a certain distance.
The formula for electric potential due to a point charge is given by:
V = k * q / r
Where:
V = Electric potential (in volts)
k = Coulomb's constant (approximately 9 x 10^9 Nm^2/C^2)
q = Charge of the point charge (in Coulombs)
r = Distance from the point charge (in meters)
In this case, we have two charges (+40x10^-6C and -60x10^-6C) placed 50 cm (or 0.5 m) apart.
To form a zero resultant potential, the magnitude of the electric potential due to the +40x10^-6C charge should be equal to the magnitude of the electric potential due to the -60x10^-6C charge.
Let's assume the distance from the +40x10^-6C charge along the line joining the two charges is 'd'.
So, the electric potential due to the +40x10^-6C charge is given by:
V1 = k * (40x10^-6) / d
And the electric potential due to the -60x10^-6C charge is given by:
V2 = k * (-60x10^-6) / (0.5 - d)
For the resultant potential to be zero, V1 should be equal to V2.
Therefore, we have the equation:
k * (40x10^-6) / d = k * (-60x10^-6) / (0.5 - d)
Now, we can solve this equation to find the value of 'd' which represents the distance from the +40x10^-6C charge along the line joining the two charges where the resultant potential is zero.