There is an isolated point charge of q=+15 mC in a vacuum at the left in Figure 18.19a. Using a test charge of q0=+0.80 mC, determine the electric field at point P, which is 0.20 m away.
| Figure 18.19
(a) At location P, a positive test charge q0 experiences a repulsive force F due to the positive point charge q. (b) At P the electric field E is directed to the right. (c) If the charge q were negative rather than positive, the electric field would have the same magnitude as in (b) but point to the left. |
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Reasoning
Following the definition of the electric field, we place the test charge q0 at point P, determine the force acting on the test charge, and then divide the force by the test charge.
Solution
Coulomb’s law (Equation 18.1), gives the magnitude of the force:
Equation 18.2 gives the magnitude of the electric field:
The electric field E points in the same direction as the force F on the positive test charge. Since the test charge experiences a force of repulsion directed to the right, the electric field vector also points to the right, as Figure 18.19b shows.
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