Two electric charges of 9 microcoulomb and minus 3 microcoulomb are pl...
Problem Statement: Two electric charges of 9 microcoulomb and minus 3 microcoulomb are placed 0.16 metre apart in a there will be a point P at which electric potential is zero on the line joining of two charges and in between them the distance of P from 9 microcoulomb charge is? Explain in details.
Explanation:
In order to find the distance of point P from the 9 microcoulomb charge, we need to follow the steps given below:
- Calculate the electric field at point P due to both the charges.
- Use the formula for electric potential to calculate the potential at point P due to both the charges.
- Set the potential at point P to zero and solve for the distance of point P from the 9 microcoulomb charge.
Calculation:
The formula for electric field at a point due to a point charge is given by:
E = k*q/r^2
where:
- E is the electric field
- k is the Coulomb constant (9 x 10^9 Nm^2/C^2)
- q is the charge
- r is the distance between the point charge and the point where electric field is to be calculated
Using the above formula, we can calculate the electric field at point P due to both the charges:
- Electric field due to 9 microcoulomb charge:
E1 = k*q1/r1^2 = (9 x 10^9)*(9 x 10^-6)/(r1)^2
- Electric field due to -3 microcoulomb charge:
E2 = k*q2/r2^2 = (9 x 10^9)*(-3 x 10^-6)/(0.16 - r2)^2
The total electric field at point P is given by:
E = E1 + E2
The formula for electric potential at a point due to a point charge is given by:
V = k*q/r
Using the above formula, we can calculate the potential at point P due to both the charges:
- Potential due to 9 microcoulomb charge:
V1 = k*q1/r1 = (9 x 10^9)*(9 x 10^-6)/r1
- Potential due to -3 microcoulomb charge:
V2 = k*q2/r2 = (9 x 10^9)*(-3 x 10^-6)/(0.16 - r2)
The total potential at point P is given by:
V = V1 + V2
Setting the potential at point P to zero, we get:
V1 + V2 = 0