The charges on three identical metal spheres are –12 mC, +4.0 mC, and +2.0 mC. The spheres are brought together so they simultaneously touch each other. They are then separated and placed on the x and y axes, as in Figure 18.40a. What is the net force (magnitude and direction) exerted on the sphere at the origin? Treat the spheres as if they were particles.

Figure 18.40  (a) Three equal charges lie on the x and y axes. (b) The net force exerted on q 1 by the other two charges is F.

Concept Questions and Answers Is the net charge on the system comprised of the three spheres the same before and after touching?

Answer   Yes. The conservation of electric charge states that, during any process, the net electric charge of an isolated system remains constant (is conserved). Therefore, the net charge on the three spheres before they touch (–12.0 mC+4.0 mC+2.0 mC=–6.0 mC) is the same as the net charge after they touch.

After the spheres touch and are separated, do they have identical charges?

Answer   Yes. Since the spheres are identical, the net charge (–6.0 mC) distributes itself equally over the three spheres. After they are separated, each has one-third of the net charge: .

Do q₂ and q₃ exert forces of equal magnitude on q₁?

Answer   Yes. The charges q2 and q3 have equal magnitudes and are the same distance from q1. According to Coulomb’s law, then, they exert forces of equal magnitude on q1.

Is the magnitude of the net force exerted on q₁ equal to 2F, where F is the magnitude of the force that either q₂ or q₃ exerts on q₁?

Answer   No. Although the two forces that act on q1 have equal magnitudes, they have different directions. The forces are repulsive forces, since all of the charges in part a of the drawing are identical. Figure 18.40b shows the force F12 exerted on q1 by q2 and the force F13 exerted on q1 by q3. To obtain the net force F, we must take these directions into account by using vector addition.

Problem solving insight Often charge magnitudes are specified in microcoulombs (mC). When using Coulomb’s law, be sure to convert microcoulombs into coulombs (1 m=10–6 C) before substituting for the charge magnitudes |q1| and |q2|.

Solution The magnitude F₁₂ of the force exerted on q₁ by q₂ is given by Coulomb’s law, Equation 18.1, as

Two point charges are lying on the y axis in Figure 18.41a: q₁=–4.00 mC and q₂=+4.00 mC. They are equidistant from the point P, which lies on the x axis. (a) What is the net electric field at P? (b) A small object of charge q₀=+8.00 mC and mass m=1.20 g is placed at P. When it is released, what is its acceleration?

Figure 18.41  (a) Two charges q 1 and q 2 produce an electric field at the point P. (b) The electric fields E 1 and E 2 add to give the net electric field E.

Concept Questions and Answers There is no charge at P in part (a). Is there an electric field at P?

Answer   Yes. An electric field is produced by the charges q1 and q2, and it exists throughout the entire region that surrounds them. If a test charge were placed at this point, it would experience a force due to the electric field. The force would be the product of the charge and the electric field.

The charge q₁ produces an electric field at the point P. What is the direction of this field?

Answer   The electric field created by a charge always points away from a positive charge and toward a negative charge. Since q1 is negative, the electric field E1 points toward it (see Figure 18.41b).

What is the direction of the electric field produced by q₂ at P?

Answer   Since q2 is positive, the electric field E2 that it produces points away from q2, as shown in the drawing.

Is the magnitude of the net electric field equal to E₁+E₂, where E₁ and E₂ are the magnitudes of the electric fields produced by q₁ and q₂?

Answer   No, because the electric fields have different directions. We must add the individual fields as vectors to obtain the net electric field. Only then can we determine its magnitude.

Solution

(a)	The magnitude of the electric fields that q1 and q2 produce at P are given by Equation 18.3, where the distances are specified in the drawing: The x and y components of these fields and the total field E are given in the following table:  Electric field   x component   y component   E1         E2         E        The net electric field E has only a component along the +y axis. So,
(b)	According to Newton’s second law, Equation 4.2, the acceleration a of an object placed at this point is equal to the net force acting on it divided by its mass. The net force F is the product of the charge and the net electric field, F=q0E, as indicated by Equation 18.2. Thus, the acceleration is