Find positions on the x axis for the charges Q1 = -1 C and Q2 = +3 C so that the electric field is zero at x = 0.
To find positions on the x-axis for the charges Q1 = -1 C and Q2 = +3 C where the electric field is zero at x = 0, we can use the principle of superposition for electric fields. According to this principle, the total electric field at any point is the vector sum of the electric fields produced by each individual charge.
Let's assume that Q1 is located at position x1 on the x-axis and Q2 is located at position x2. We need to determine the values of x1 and x2 such that the total electric field is zero at x = 0.
The electric field produced by a point charge Q at a distance r is given by Coulomb's Law:
E = k * Q / r^2
where k is the electrostatic constant (k = 8.99 × 10^9 Nm^2/C^2).
For Q1 at position x1, the electric field is directed towards Q1 if Q1 is negative. So the electric field at x = 0 due to Q1 is given by:
E1 = k * Q1 / (x1 - 0)^2 [Since the distance between Q1 and x = 0 is x1 - 0 = x1]
For Q2 at position x2, the electric field is directed away from Q2 if Q2 is positive. So the electric field at x = 0 due to Q2 is given by:
E2 = k * Q2 / (0 - x2)^2 [Since the distance between Q2 and x = 0 is 0 - x2 = -x2]
Since the total electric field at x = 0 is zero, the sum of the electric fields due to Q1 and Q2 must be zero:
E1 + E2 = 0
Substituting the expressions for E1 and E2, we get:
k * Q1 / (x1 - 0)^2 + k * Q2 / (0 - x2)^2 = 0
Simplifying the equation, we have:
Q1 / x1^2 + Q2 / x2^2 = 0
Now we can substitute the values of Q1 and Q2 (-1 C and +3 C) into the equation:
(-1 C) / x1^2 + (3 C) / x2^2 = 0
To solve the equation for x1 and x2, we need to choose appropriate values that satisfy the equation. One possible solution is x1 = 1 m and x2 = 3 m. This gives:
(-1 C) / (1 m)^2 + (3 C) / (3 m)^2 = 0,
-1 C + 1 C = 0.
Therefore, if Q1 is located at x1 = 1 m and Q2 is located at x2 = 3 m on the x-axis, the electric field will be zero at x = 0.