Two charges are D = 1.5 m apart and lie on the x-axis. What must be the magnitude of the charge on the right one, Q2, (in C) so that the electric field is zero at x = 1.05 meters. Let Q1 be at the origin and have a charge 6 C.
To find the magnitude of the charge on the right one (Q2) such that the electric field is zero at x = 1.05 meters, we can make use of the formula for the electric field due to a point charge. The electric field (E) at a point due to a charge (Q) is given by:
E = k * Q / r^2
where k is the electrostatic constant, Q is the magnitude of the charge, and r is the distance from the charge to the point.
In this case, we have two charges, one at the origin (Q1) with a charge of 6 C, and the other on the x-axis (Q2) at a distance of 1.5 m from Q1. We need to find the value of Q2.
To determine the electric field at x = 1.05 meters, we need to consider the electric field due to both charges. The electric field at a point due to multiple charges is the vector sum of the individual electric fields due to each charge.
Since we want the electric field to be zero at x = 1.05 meters, the sum of the electric fields due to Q1 and Q2 should be zero at that point.
Let's calculate the contribution of Q1 to the electric field at x = 1.05 m:
E1 = k * Q1 / r^2
= (9 * 10^9 Nm^2/C^2) * (6 C) / (1.05 m)^2
Now, let's calculate the contribution of Q2 to the electric field at x = 1.05 m:
E2 = k * Q2 / r^2
= (9 * 10^9 Nm^2/C^2) * (Q2 C) / (1.05 m + 1.5 m)^2
Since the total electric field is the sum of E1 and E2, we can set their sum to zero and solve for Q2:
E1 + E2 = 0
Substituting the calculated values of E1 and E2, we have:
(9 * 10^9 Nm^2/C^2) * (6 C) / (1.05 m)^2 + (9 * 10^9 Nm^2/C^2) * (Q2 C) / (1.05 m + 1.5 m)^2 = 0
Now, we can solve this equation to find the value of Q2.
By rearranging and simplifying the equation, we have:
(6 / 1.05^2) + (Q2 / 2.55^2) = 0
Multiply both sides by (1.05^2 * 2.55^2), we get:
6 * 2.55^2 + Q2 * 1.05^2 = 0
Simplifying further:
Q2 = - (6 * 2.55^2) / 1.05^2
Calculating this expression will give us the magnitude of the charge Q2 in coulombs (C) that will result in a zero electric field at x = 1.05 meters.