The electric field at the point x=5.00cm and y=0 points in the positive X direct

Question

The electric field at the point x=5.00cm and y=0 points in the positive X direction with a magnitude of 10.0 N/C. At the point X=10.0 cm and y=0 the electric field points in the positive X direction with a magnitude of 17.0 N/C . Assume this electric field is produced by a single point charge.

Part A

Find the charges location

Part B

Find the magnitude of the charge

Please answer both questions and explain because I tried to figure it out on my own using other chegg study questions and all the explinations were so confusing your help is appreciated thank you

this is due tomorrow.  

Explanation / Answer

Start with F = qE and F = kqq/d2

Thus E = kq/r2 and solving for q provides q = Er2/k

Since q is the same charge, two formulas can be set equal using the two differect electric fields. Additionally it is known that the electric field points in the positive x direction at both the 5 cm and 10 cm locations. It is also shown that the Electric Field is larger at the 10 cm location which provides for the fact that the charge is on the positive x-axis beyond the 10 cm mark so that it is closer to the 10 cm point than it is to the 5 cm point. Therefore setting the equations equal yields...

Part A

E1(x-5)2/k = E2(x-10)2/k and k can be eliminated since it is the same on both sides of the equation.

10(x-5)2 = 17(x-10)2

Take the square root of both sides

10 (x-5) = 17 (x-10) or 3.16(x-5) = 4.123(x-10)

Distribute = 3.16x - 15.81 = 4.123x - 41.23

Solving for x provides 25.42 = 0.96x or x = 26.47 cm

(Note that it was not necessary for part A to convert the distance to meters since a ratio was used, however in part B, it will be necessary)

Part B...

Substitute the x location into either of the formulas used in part A. I will choose

q = E1(.2647 - .05)2/k = 10(.2647 - .05)2/9 X 109 = 5.12 x 10-11 C

to prove it correct, also substitute into the second formula

q = E2(.2647 - .1)2/k =17(.2647 - .1)2/9 X 109 = 5.12 x 10-11 C

In case it is needed, using the fact that the Electric Field points in the direction of the positive X axis, and therefore points toward this charge, the charge itself is negative by the definition of Electric Field

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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