Two equal negative point charges q1 = q2 = -q are respectively located at points (0,a) and (0,-a). A positive point charge q0 of mass m moves from point B (-b,0) to the origin of the Cartesian coordinate system with an initial velocity v0.
- Draw the net electric field E due to the charges q1 and q2 at point B and the force vector exerted on q0 when it is located at this point.
- Calculate the electrostatic potential due to the charges q1 and q2 both at B and the origin of the Cartesian coordinate system.
- Calculate the velocity of q0 when it reaches the origin of the Cartesian coordinate system.
- Calculate the distance traveled by q0 from the origin of the Cartesian coordinate system before it comes to a rest.
Givens: k = 1/(4πε0); ε0 = 8.854 10-12 C2/(N·m2)
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Solution:
The situation described in the problem statement is represented in the following figure:
First we are going to represent the net electric field due to the charges q1 and q2 at point B to visualise the force exerted on q0. We will name E1 the field due to the charge q1 and E2 the field due to the charge q2.
To find the direction of the vector E1 we perform a “thought experiment” that consists in placing a positive test charge at point B and to find out the direction of the force it would experience in presence of charge q1. Since q1 is negative the test charge would be attracted, and therefore E1 points towards q1. Recall that negative charges are sinks of electric field lines. We find the direction of the vector E2 by performing the same “thought experiment” for q2.
The net field E is the sum of both vectors and we can determine it graphically with the help of the parallelogram law. The three vectors are represented on the left side of the next figure.
If we place the charge q0 at point B, the force exerted on it is given by:
And, as q0 is positive, the vector F is parallel to E, as seen on the right side of the previous figure.
The electric potential at B is the sum of the potentials due to q1 and q2 at this point:
Where r is the distance from each charge to point B. We use Pythagoras’ theorem to find r1 and r2:
And after substituting in the potential expression we obtain:
We use the same method to calculate the electric potential at the origin of the Cartesian coordinate system O:
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The electrostatic force is conservative, therefore the net energy of charge q0 is the same at both points B and O. If we name K the kinetic energy and v the velocity of the charge at point O, the sum of the kinetic energy and the electrostatic potential energy is constant:
After substituting the potential and isolating v2 we obtain:
The charge q0 will come to a rest at point C of coordinates (c,0) when its velocity becomes zero. The potential at point C is:
And after applying the conservation of energy between point B and C, we get:
Finally, after substituting the potential at both points we obtain:
And we can isolate the value of c from the previous expression.
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