Question

B. At one point along a line there is an electric field pointing to the left. The source of this field is one point charge somewhere on that line. x= 10 cm E| = 800 N/C. 1. Which of the following possibilities could have caused that field? (More than one may be correct.) a. A positive charge to the right of the point. b. A positive charge to the left of the point. c. A negative charge to the right of the point. d. A negative charge to the left of the point. 2. After making more measurements, we find the magnitude and direction of the field at another point: x= 10 cm x= 20 cm E = 200NC E = 800 NC 3. Now, only one of the possibilities listed in part I can be correct. Which one is it? 4. Look at the factor by which the field strength changes when moving from x=20 cm to x = 10 cm. That is the piece of information that tells you where the source charge is. It is at x=_ __cm. c. Still thinking about the situation in Question B. After answering part 5 of Question B, you have enough information to calculate the value of the source charge (source of the electric field). Do that calculation. Show your calculation.

Answer 1

x = 10cm El=goonk (1 nie ОА And know that the direction of electroc field due to positive change from it. saway. the direction of electric field due to a negative change as towands it, so, the following tovo papabilities are ocenenio , (a) a positive change to the right negative change to the left of the pond. possible for the above care they of the point. x=10cm 9= 20 cm 6 1E1 = 200 alle El=&conlle Cue Socrace of electric field as the electric field ♡ We know that the mag ni aiteele of electore held decreases when from the point change/ and the increases towards the source magnitude change cwe can see that kigune so, from the above xa10cm to HERO acej nitude of increases as si so, the only possibility (a) A poortive change to the right of the paint. 6 Let the charge and meegnstudes a oes shown bigure below cm go ben is at aay goeroce x=ojem 9= 0. Qr) Ela 20onk El- 800 olle

The distance between source choonge anel the point x=0,9 m 3 (x-0.21), so, the electric field in given by (E) = ko (% -0,2)? (E) CX-0.2)? k co Again Q = (Rou) (X-0.2)? [urli euhere k=9x109 Nm2 k Again the distance belonen monte (x-0,1m) ,30, electric field charge and the point Malolm å given by KO (X-0.12 Q- El (x-o.)? (200) (x-o.,02 K k equation As the same source change has been usead so equesting and @ (800) (x-0,2)² (20) (x-0.1)? 20 800 Taking 2 (x-0.2) (x-0.10² square root on both sides X-0.2 g-0.) 2x -0.4 X-On X = -0.1 +0.4. 0,3 08 x= 0.3 m cor) x=30 cm