A single point charge is located at an unknown point on the x-axis. There are no

Question

A single point charge is located at an 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 5200 N/C, directed in the +x direction, while the field at x = +5.0 m has a magnitude of 1300 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 C and the charge is located at x = 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 C and the charge is located at x =

Explanation / Answer

No direction is given for the second field. We know it is further away from the charge because the value is lower.

However the two possibilites are a) the charge is located to the left of the point +2.0 m ( but still close to that point) or the charge is located to the right of 2.0 m

In each case, E is proportional to Q/d

If the charge is to the left of 2.0 m then distance to charge 1 is y , distance to charge 2 is (y + 3.0) m
and y2/y1=5200/1300= 4
so (y+3) / y =4 ==> y= 1 m (x=1.0 m from origin)

E= 9*10^9*q/1.0= 5200

q= 5.78*10^-7 C


b) If the charge is located to the right of 2.0 m then distance to charge 1 is y distance to charge to is (3.0 - y)
again y2/y1 =4
so (3.0-y)/y=4 == . y= 0.6 m    (x=2.6 m from origin)

E= 9*10^9*q/0.6= 5200

q= 3.47*10^-7 C

or x=1m and x=2.6 m respectively from origin

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.