In (Figure 1) , two objects, O1 and O2, have charges +1.0 ? C and ?3.5 ? C, resp
ID: 2002274 • Letter: I
Question
In (Figure 1) , two objects, O1 and O2, have charges
+1.0?C and ?3.5?C, respectively, and a third object, O3, is electrically neutral.
Part A : What is the electric flux through the surface A1 that encloses all three objects?
Part B : What is the electric flux through the surface A2 that encloses the third object only?
2.
PArt A;
Determine magnitude of the electric field at the point P shown in the figure(Figure 1) . The two charges are separated by a distance of 2a. Point P is on the perpendicular bisector of the line joining the charges, a distance x from the midpoint between them.
Express your answer in terms of Q, x, a, and k.
Part B: Determine the direction of the electric field. Assume that the positive x and y axes are directed to the right and upward respectively.
2Explanation / Answer
1) a) From Gauss's law: Flux = c E . dA = q/eo
Since this surface encloses all charge, we can simplify:
Flux = (q1+q2+q3)/eo
Flux = ( (1*10^-6)+(-3.5*10^-6)+(0) )/(8.85*10^-12) = -282485.88 N·m2/C
b) A2 contains no charge so the flux through it is zero. The flux entering A2 due to charges Q1 and Q2 is equal to the flux leaving A2. This has to be true by Gauss Law.
2) Let the field from +Q be E(+) and the field from -Q be E(-)
E(+) = KQ / (a^2 + x^2)
E(-) = - KQ / (a^2 + x^2)
These are the two vectors that are marked on the diagram. But the resultant field is marked as E.
To find E, we have to work out the horizontal and vertical components of E(+) and E(-). Can you see from from the symmetry that the horizontal components point in opposite directions and thus cancel?
So, we need to work out the vertical components of E(+) and E(-), which are both pointing downwards from P.
Magnitude of vertical component of E(+) is [KQ / (a^2 + x^2)] * sin
Magnitude of vertical component of E(-) is [ KQ / (a^2 + x^2)] * sin
So E = E(+) + E(-) = 2 [KQ / (a^2 + x^2)] * sin (pointing downwards from P.)