Consider two point charges: the source charge Q and the test charge q, separated by distance r.
We can say that Q is exerting an electrical force of on q, as according to Coulomb’s Law.
Alternatively, we can say that Q is creating an electrical field E around it, causing q to experience an electric force of qE.
Since electrical field strength is force per unit (positive) charge, the strength of the electric field produced by Q at distance r away is
Notice that the field strength obeys the inverse-square law.
What about the direction? Since a positive source charge will repel positive test charges radially outward, a positive source charge produces an electric field that is radially outward. Conversely, since a negative source charge will attract positive test charges radially inward, a negative source charge produces an electric field that is radially inward.
QA and QB are point charges of +2.1 pC and −2.1 pC bound at vertices A and B of an equilateral triangle of side 3.0 um as shown in the diagram below. Calculate
a) the electric field strength at C.
b) the electric force experienced by
i) an electron
ii) a proton
placed at c
c) the electric field strength midpoint between A and B.
The magnitude of the field strength of QA at C,
The magnitude of the field strength of QB at C,
Note that QA exerts a radially outward field whereas QB exerts a radially inward field, EA and EB are thus in the directions shown in the diagram. The resultant field strength at C is thus given by the vector summation of EA and EB. Note that the vector summation triangle is an equilateral triangle. So the resultant field strength at C is
Since ER is rightward, the proton experiences a rightward force whereas the electron experiences a leftward force.
At the midpoint, both EA and EB are rightward and have the same magnitude.
The resultant field strength is thus