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Chrome File Edit View History Bookmarks People Window Help 34% D Tue Feb 2 20:15 Katrina Catalano QA E Katrina CIn The Figure Above x MG If you move from or x PY 106 (24 unread) X Bb My courses Dashbo X Hi! l Can't seem To x VYf the Electric Field x I PY106 Spring 201 C www.webassign.net/web/Student/Assignment-Responses/submit?dep 13285688 Additional Materials eBook 3. G+ 3/4 points I Previous Answers BUElemPhys 6. P.059 My Notes Ask Your Teacher A single point charge is located at a unknown point on the x-axis. There are no other charged objects nearby. You measure the electric field at x 2.0 m to be 4800 N/C, directed in the +x direction, while the field at x 5.0 m has a magnitude of 1200 N/C. What is the sign and magnitude of the point charge? There are two possible solutions. First, we'll do the solution in which the charge on the point charge has the larger magnitude For this solution, the sign and magnitude of the charge is 4.804e-6 C and the charge is located at x -1 Now, we'll do the solution in which the charge on the point charge has the smaller magnitude. For this solution, the sign and magnitude of the charge is -1.201e-6 X C and the charge is located at x 3 Additional Materials eBook 4. G+ 2/2 points I Previous Answers BUElemPhys 6. P.103 My Notes C+ Ask Your Teacher Eight different charged balls, which we treat as point charges, are arranged in a highly symmetric pattern around a square. Note that the value of Q is 3.00 x 10 o C, and

Explanation / Answer

let's say the charge is at a distance d from the origin

then Electric field due to this charge at x = +2 is

ke*q/x^2 = 4800

(9*10^9*q)/(d-2)^2 = 4800

charge is q = 4800*(d-2)^2/(9*10^9)

at the x = +5 m

then q = 1200*(5-d)^2/(9*10^9)

then we can write down

4800*(d-2)^2/(9*10^9) = 1200*(5-d)^2/(9*10^9)

4*(d-2)^2 = (5-d)^2

2*(d-2) = (5-d)

2d-4= 5-d

3d = 9

d = 3 m

then q = (4800*(3-2)^2)/(9*10^9) = 0.5333*10^-6 C

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